Game making expression positive definite

On the board there are six numbers $A,B,C,D,E,F$, all initially zero. Mary and Nancy are playing a game as follows. On Mary's turn, she increases one of $A,B,C$ by 1, and on Nancy's turn, she increases one of $D,E,F$ by 1. They take turns alternatingly until each has gone 2019 times. Each pair of turns is called a "set" (for example if Mary moves first, then 1 set consists of Mary's move followed by Nancy's move).

Winning condition: Mary wins if at any point at the end of a set the following inequality is true for all $x,y$ real numbers: $$Ax^2 + By^2 + C \geq Dxy + Ex + Fy$$ Determine who has the winning strategy if:

1. Mary goes first

2. Nancy goes first

Solution

If Mary goes first Nancy has a winning strategy.

Replace $x$ with $x/z$ and $y$ with $y/z$ to make the inequality homogeneous: $$Ax^2 + By^2 + Cz^2 \geq Dxy + Exz + Fyz$$ Note that the following form is always true if $u,v,w \geq 0$ $$u(x-y)^2 + v(x-z)^2 + w(y-z)^2 \geq 0$$ So if at any point after a set the inequality can be reduced to that form, then Mary wins. We claim the following two things:

1. That Nancy can always avoid that form, and

2. That if the form is not achieved then there exists a $x,y,z$ to make the inequality false

First to prove 1, note the following rearrangement: $$u(x-y)^2 + v(x-z)^2 + w(y-z)^2 \geq 0$$ $$(u+v) x^2 + (u+w)y^2 + (v+w)z^2 \geq 2u xy + 2v xz + 2w yz$$ Therefore if $D = (A+B-C), E = (A+C-B), F = (B+C-A)$ then that form is achieved. This is the unique solution to achieve that form, and that solution may be negative. If the solution to that form contains negative number then no matter what Nancy does, that form is not achieved. But even if it is a triple of non-negative integers, Nancy has 3 choices of moves to make, so she still has at least 2 moves that won't result in that form.

Now to prove 2, we show that we can find $x,y,z$ that violates the inequality.

First, if $A,B,C$ do not form a triangle, then $A>B+C$ (or its permutation). Note that if we assume (WLOG) that $A>B+C$ then $A+B>C,A+C>B$. Furthermore WLOG we may assume that $B \geq C$. $$Ax^2 + By^2 + Cz^2 \geq Dxy + Exz + Fyz$$ $$\iff (A+B-C)(x-y)^2 + (A+C-B)(y-z)^2 + (B+C-A)(x-z)^2 \geq -(2A+2B-2C-2D)xy - (2A+2C-2B-2E)yz - (2B+2C-2A-2F)xz$$ In that case we can choose the following variables. Let $S,R$ be large numbers. Let $x= 1/R, y = R + 1/R, z = SR + 1/R$. Then $$ (A+B-C)R^2 + (A+C-B)(S-1)^2R^2 \geq (A-B-C)S^2R^2 + ...$$ The "..." part in RHS consists of terms $SR^2$ or lower. As we let $S,R$ become large, the dominant terms are $(S-1)^2R^2$ versus $S^2R^2$. For large enough $S$, $(A+B-C)+(A+B-C)(S-1)^2 < (A-B-C)S^2$. At that point we just fix $S$ but let $R$ become larger. Since the coefficient of $R^2$ in the RHS is larger, then the RHS grows faster, invalidating the inequality.

Now if $A,B,C$ form a triangle, $$\iff (A+B-C)(x-y)^2 + (A+C-B)(y-z)^2 + (B+C-A)(x-z)^2 \geq (2A+2B-2C+2D)xy + (2A+2C-2B+2E)yz + (2B+2C-2A+2F)xz$$ Note that the sum of coefficients in RHS is zero, so the terms are not all positive. We shall divide into two cases: one positive two negatives or vice versa

Case 1: if the RHS is of the form $Uxy + Vxz - (U+V) yz$ with $U,V \geq 0$. Then $$RHS = Uy(x-z) + Vz(x-y)$$ Then we set $x = R+1/R, y = R, z= R$. The terms in LHS will be zero or $(1/R^2)$ whereas the terms in RHS will be $(U + V)$. By setting $R$ large enough we can make LHS < RHS.

Case 2: if the RHS is of the form $(U+V)xy - Uyz - Vxz$ with $U,V \geq 0$ Then $$RHS = Uy(x-z) + Vx(y-z)$$ Then we set $x = R+1/R, y = R+1/R, z= R$. The terms in LHS will be zero or $(1/R^2)$ whereas the terms in RHS will be $(U + V)(1+1/R)$. Again, by setting $R$ large enough we can make LHS < RHS.

Challenge Can you generalize this to third power and four variables? In other words, can the game be extended into the following inequality? $$Ax^3 + By^3 + Cz^3 + Dw^3 \geq Exyz + Fxyw + Gxzw + H yzw$$ Answer: yes. We make use of the following property: $f(a,b,c) = a^3 + b^3 + c^3 - 3abc = (a+b+c)((a-b)^2 + (b-c)^2 + (c-a)^2)/2 \geq 0$ If the inequality can be expressed as a positive sum of $f(x,y,z), f(x,y,w), f(x,z,w), f(y,zw)$ then the inequality holds. If not, then there are two cases:

1. One or more of the $f$ form occurs on the RHS, in which case we choose $x,y,z,w$ to maximize the growth of the ones in the RHS. For example, $x=y=1/R, z = R+1/R, w = SR + 1/R$ for large $S,R$.

2. The $f$ forms all occur on the LHS, so there are terms of $xyz$ etc on the RHS whose coefficients all sum to zero. We can pick $x,y,z,w$ whose differences are small but themselves are large, such as $x=R, Y = R+1/R, z = R+2/R, w = R+3/R$. That way, the values of $f$ will be small but values of $xyz$ will be big. By judiciously permuting those values depending on which coefficients are negative, we can make LHS to be < RHS.

Maximum point above triangle

In triangle $ABC$, $AD$ is an internal bisector. Choose point $X$ on $DA$'s extension, and let $Y,Z$ be on $XB$ and $XC$ so that $AY \perp XB$ and $AZ \perp XC$. Define $f(X)$ as: $$f(X) = \frac{AY}{XB} + \frac{AZ}{XC}$$ Find the point $X*$ along $DA$'s extension such that $f(X*)$ is maximum.

Solution

Let $\theta = \angle XAB = \angle XAC$. Easy to see that $\pi/2 < \theta < \pi$. Let $x = AX$.

Area of $\triangle ABX$ = $$\frac{1}{2}cx \sin \theta = \frac{1}{2} XB.AY$$ So: $$\frac{AY}{XB} = \frac{cx \sin \theta}{XB^2} = \frac{cx\sin \theta}{c^2+x^2-2cx \cos \theta}$$ And likewise: $$\frac{AZ}{XC} = \frac{bx\sin \theta}{b^2+x^2-2bx \cos \theta}$$ Now suppose WLOG $c > b$ and $c = tb$ (with $t > 1$). Then we need to maximize: $$\frac{cx}{c^2+x^2-2cx \cos \theta} + \frac{bx}{b^2+x^2-2bx \cos \theta}$$ $$ \frac{\frac{x}{bt}}{(\frac{x}{bt})^2 + 1 - 2 t\frac{x}{bt}\cos \theta }+\frac{\frac{x}{b}}{(\frac{x}{b})^2 + 1 - 2 \frac{x}{b}\cos \theta }$$ So if we let $y = x/b$, and $r = -2 \cos \theta$ (which means $0 < r < 2$), and $g(s) = s/(s^2+rs+1)$ then the problem is akin to finding the minimum of the following function for $y > 0$: $$h(y) = g(y) + g(\frac{y}{t})$$

For $x,y,z$ positive numbers such that $xyz = 1$, prove that: $$2 \sqrt{2} (x^7+y^7)(x^7+z^7)(y^7+z^7) \geq \sqrt{(x^{16}+7)(y^{16}+7)(z^{16}+7)}$$

Solution

$$2(x^7+y^7)(x^7+z^7) = x^{14} + (x^{14} + 2x^7(y^7+z^7) + 2y^7z^7)$$ by AM-GM: $$ \geq x^{14} + 7x^6y^4z^4 = \frac{x^{16}+7}{x^2}$$ By multiplying similar inequalities and taking square root, we get the desired result

Find all m for inequality

Find all real number $m$ such that, for all positive real numbers $x,y,z$ the following is true: $$ \frac{x}{x^2 + myz} + \frac{y}{y^2 + mxz} + \frac{z}{z^2 + mxy} \geq \frac{9}{(m+1)(x+y+z)}$$

Solution

First we show that for $m \geq 8$ it's true. By Cauchy: $$\frac{x}{x^2 + myz} + \frac{y}{y^2 + mxz} + \frac{z}{z^2 + mxy} \geq \frac{(x+y+z)^2}{x^3 + y^3 + z^3 + 3m.xyz}$$ Then let $A = x^2 + y^2 + z^2,B = xy+yz+zx$. Because $x^3+y^3+z^3 = (x+y+z)(A -B)+3xyz$, $$ = \frac{A+2B}{(x+y+z)(A-B)+3(m+1)xyz} \geq \frac{9}{(m+1)(x+y+z)}$$ The last inequality is equivalent to (because $A \geq B$ we may multiply by the denominators without changing the sign): $$(A+2B)(x+y+z)(m+1) \geq 9(x+y+z)(A-B) + 27(m+1)(x+y+z)$$ $$\iff (x+y+z)[(m-8)A + (2m+11)B] \geq 27(m+1)xyz$$ Indeed by AM-GM: $$x+y+z \geq 3 \sqrt[3]{xyz}$$ $$ A \geq 3 \sqrt[3]{x^2y^2z^2}$$ $$ B \geq 3 \sqrt[3]{x^2y^2z^2}$$ So: $$ (x+y+z)[(m-8)A + (2m+11)B] \geq 3 \sqrt[3]{xyz} [3(m-8) \sqrt[3]{x^2y^2z^2} + 3(2m+11)\sqrt[3]{x^2y^2z^2} = 27(m+1)xyz$$ Now to show necessity, plug in $z = t, x=y=1$, then the left hand side is : $$ \frac{2}{1+mt} + \frac{t}{t^2+m} \geq \frac{9}{(m+1)(2+t)}$$ $$(2(t^2+m)+t(mt+1))(2+t)(m+1) \geq 9(t^2+m)(mt+1)$$ which has to be true for all $t \geq 0$. However, the coefficient of $t^3$ on the LHS is $m(m+1)$ and on the RHS is $9m$. Because this inequality has to be true for all $t$ no matter how big, then the coefficient on the LHS has to be greater than or equal to that of the RHS, otherwise we can choose $t$ large enough to violate that expression. $$m(m+1) \geq 9m$$ which means $m \geq 8$.

Inequality a b c

For $a,b,c \geq 0$ prove that: $$2(a+b+c)^2 + 3(ab+bc+ca) \geq (a+b+c)(\sqrt{a} + \sqrt{b} + \sqrt{c})^2$$

Solution

This is equivalent to: $$(a+b+c)^2 + 3(ab+bc+ca) \geq 6(a+b+c)(\sqrt{ab} + \sqrt{bc} + \sqrt{ca})$$ which is the cyclic sum of: $$(a+b+c)^2 + 9ab \geq 6 \sqrt{ab}(a+b+c)$$ which is true by AM-GM.

Generalization

Find all $\lambda$ such that this holds for all $a,b,c \geq 0$: $$(a+b+c)^2 + \lambda (ab+bc+ca) \geq \frac{\lambda + 3}{9}(a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c})^2$$

Solution

For $0 \leq \lambda \leq 3/2$ we can show it by proving the original problem, and realizing that: $$3(a+b+c)^2 \geq (a+b+c)(\sqrt{a} + \sqrt{b} + \sqrt{c})^2$$ is just AM-RMS.

Inequality x y z

If $x,y,z$ are positive numbers such that $x+y+z = 1$ show that: $$ \frac{1}{x+y} + \frac{1}{x+z}+ \frac{1}{y+z} + \frac{9}{2} \geq \frac{3}{1 - (\frac{x-y}{2})^2} + \frac{3}{1 - (\frac{y-z}{2})^2} + \frac{3}{1 - (\frac{x-z}{2})^2}$$

(Hopefully) Correct Solution

WLOG we may assume that $x \geq y \geq z$. And for now we assume that $2y \geq x+z$ (the case where $2y < x+z$ is handled later below).

By AM-HM: $$\frac{1}{y+z} + \frac{1}{y+z} + \frac{1}{x+z} \geq \frac{9}{(y+z)+(y+z)+(x+z)} = \frac{9}{1+y+2z} = \frac{9}{2+z-x}$$ $$\frac{1}{y+z} + \frac{1}{x+z} + \frac{1}{x+z} \geq \frac{9}{1+x+2z} = \frac{9}{2+z-y}$$ Adding the two inequalities: $$ \frac{1}{y+z} + \frac{1}{x+z} \geq \frac{3}{2+z-x} + \frac{3}{2+z-y} $$ (Call this inequality 1).

Incorrect Solution

WLOG we may assume that $x \geq y \geq z$. Suppose $a = x-z, b = x-y$ so $a+b = 2x-y-z = 3x-1$. If $a=b=0$ then $x=y=z=1/3$ and equality happens. Thus at least one of $a,b$ is positive, so $a+b > 0$. Now, by Cauchy: $$(\frac{a}{a+b} . \frac{1}{y+z} + \frac{b}{a+b} . \frac{3}{2})(\frac{a(y+z)}{a+b} + \frac{2b}{3(a+b)}) \geq (\frac{a}{a+b} + \frac{b}{a+b})^2 = 1$$ So: $$\frac{a}{a+b} . \frac{1}{y+z} + \frac{b}{a+b} . \frac{3}{2} \geq \frac{3(a+b)}{3a(y+z) + 2b} = \frac{3}{2+z-x}$$ The last equality is equivalent to: $$ \frac{1(a+b)}{3a(y+z) + 2b} = \frac{1}{2+z-x} $$ $$ \iff (a+b)(2+z-x) = 3a(y+z) + 2b $$ Because $2+z-x \iff 3-2x-y$ and $y+z = 1-x$ $$ \iff (3x-1)(3-2x-y) = 3(x-z)(1-z) + 2(x-y)$$ Upon expanding and rearranging: $$ \iff 0 = 3(x-1)(x+y+z - 1) $$ which is true. On the other hand, using the same definition of $a,b$, we can also show: $$\frac{b}{a+b} . \frac{1}{y+z} + \frac{a}{a+b} . \frac{3}{2} \geq \frac{3(a+b)}{3b(y+z) + 2a} = \frac{3}{2+y-x}$$ (Using similar identity as above) Adding the two inequalities, we have: $$ \frac{1}{y+z} + \frac{3}{2} \geq \frac{3}{2+z-x} + \frac{3}{2+y-x}$$ And similarly: $$ \frac{1}{x+z} + \frac{3}{2} \geq \frac{3}{2+x-y} + \frac{3}{2+z-y}$$ $$ \frac{1}{x+y} + \frac{3}{2} \geq \frac{3}{2+y-z} + \frac{3}{2+z-x}$$ Now all we need to show is that the sum of LHS is equal to the LHS of the given problem. Indeed it is so because: $$ \frac{3}{2+y-x} + \frac{3}{2+x-y} = \frac{3(2+x-y) + 3(2+y-x)}{2^2 - (x-y)^2} = \frac{12}{4 - (x-y)^2} = \frac{3}{1 - (\frac{x-y}{2})^2}$$ Edit: what is wrong with this proof?

Inequality a,b,c

For $a,b,c$ non-negative real numbers such that $a+b+c=1$, prove that: $$(3a+1)(3b+1)(3c+1) \geq 3 \sqrt{3}(\sqrt{a} + \sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})$$

Solution

With Cauchy we have: $$(3a + 1)(1 + 3b) \geq (\sqrt{3a} + \sqrt{3b})^2 = 3(\sqrt{a} + \sqrt{b})^2$$ Multiplying all of the similar inequalities, we get the desired result. Equality happens if and only if $3a = 1/(3b), 3b = 1/(3c), 3c = 1/(3a)$, which means $a=b=c = 1/3$.

maximum minimum of function

For $A,B,C$ non-negative angles such that $A+B+C = \pi / 2$, find the maximum and minimum of: $$ f = \sin A + \sin B + \sin C + \sin^2 A + \sin^2 B + \sin^2 C$$ and $$g = \cos A (\sin A -1) + \cos B (\sin B - 1) + \cos C (\sin C - 1)$$

AM-GM-HM

From OSP SMA 2018 For positive $a,b,c$ such that $1/a + 1/b + 1/c = 3$ prove that: $$a+b+c + \frac{4}{1+(abc)^\frac{2}{3}} \geq 5$$

Inverse distances to points

Let $P$ and $Q$ be two distinct points on a plane. For any given point $X \neq P,Q$, define two functions $f(X)$ and $g(X)$ as follows: $$f(X) = \frac{1}{PX} + \frac{2}{QX}$$ $$g(X) = \frac{2}{PX} + \frac{1}{QX}$$ Now for any positive number $t$, let $S(t)$ be the set of all points $Y$ such that $f(Y) \leq tg(Y)$.

Prove that $S(t)$ is bounded

Prove that $S(t)$ is convex. That is, if $A,B$ are in $S(t)$ then $AB$ is in $S(t)$.

Identify all points $X$ on $S(t)$ such that $PX+QX$ is minimum.

Prove that the area of $S(t)$ is a convex function of $t$.

Powered distances

Let $p$ and $q$ be two infinitely long non-parallel lines on a plane. For any point $X$, let $d_p(X), d_q(X)$ denote distances from $X$ to $p$ and $q$ respectively. For any positive number $r$, let $S(r)$ be the set of all points $Y$ such that $d_p(Y)^{2015} + d_q(Y)^{2015} \leq r$.

Prove that $S(r)$ is bounded.

Prove that $S(r)$ is convex. That is, if $A$ and $B$ are in $S(r)$ then $AB$ is in $S(r)$.

Identify all points in $S$ such that $d_p(X)^{2014} + d_q(X)^{2014}$ is maximum.

Prove that the area of $S(r)$ is a convex function of $r$.

Fractional inequality

Let $x, y > 0$, $x \neq y$, and let $n > 1$ be an integer.

If $x^n - y^n = x^{n+1} - y^{n+1}$, show that $$1 < x+y < \frac{2n}{n+1}$$

Solution Let $f(x) = x^n ( 1 - x)$ and we have $f(x) = f(y)$. WLOG, we may assume that $x If $y > 1$ then $f(y)$ is negative, so that $f(x)$ is negative, which means $x > 1$ as well. But $f(x)$ is monotonically decreasing for $x > 1$ (because $x^n$ is monotonically increasing and $1-x$ is monotonically decreasing). Contradiction.

If $y=1$ then $f(x) = f(y) = 0$ impossible for $x>0$. Contradiction.

If $0 < x < y < 1$, we draw the graph of $f(x)$. It starts off at (0,0), going up to a maximum between 0 and 1, and goes back down to (1,0). This graph can be deduced easily be realizing that $f(x) = x(1-x) . x^{n-1}$. Because $x(1-x)$ only has one turning point and $x^{n-1}$ is monotonically increasing (or constant), then $f(x)$ also only has one turning point.

We can find that turning point by using AM-GM: $$f(x) = x.x\dots x.(1-x)$$ $$ = \frac{1}{n} x.x\dots x.(n-nx)$$ $$ \leq \frac{1}{n} (\frac{n}{n})^n = \frac{1}{n}$$ Equality happens when $x = n-nx$ or $x = \frac{n}{n+1}$. Note that for $n>1, \frac{n}{n+1} = 1 - \frac{1}{n+1}> 1/2$

Inequality

If $x_1,\dots,x_n$ are real numbers such that $0 \leq x_i \leq 1 \forall i$, and satisfies $x_1 + \dots + x_n = m+r$ where $m$ is an integer and $0 \leq r < 1$, show that:

$$(1+x_1)(1+x_2)\dots(1+x_n) \geq (1+r)2^m$$

Determine the conditions under which equality is achieved.

Solution

For each $i$ such that $x_i = 0$ or $x_i = 1$, we can safely eliminate them from the problem without changing neither the condition nor the inequality. So without loss of generality, we can assume that $0 < x_i < 1$ for all $i$.

Now, we prove by induction on $n$. It is trivial for $n=1$. For $n=2$, we have two cases:
1. The simple case, $x_1 + x_2 = r < 1$ then the inequality also holds trivially
2. The overflowing case, where $1 < x_1 + x_2 = 1 + r$, then:
$$(1-x_1)(1-x_2) \geq 0$$
$$\iff 1+x_1x_2 \geq x_1+x_2 = 1+r$$
$$\iff 1+x_1+x_2+x_1x_2 \geq 2+2r$$
$$\iff (1+x_1)(1+x_2) \geq 2(1+r)$$

Now, for a larger value of $n$, suppose that the inequality is established for $n-1$, and suppose that $x_1 + \dots + x_{n-1} = m + r$. Again we have two cases,

1. The simple case, if $x_n + r < 1$ then $x_1 + \dots + x_n = m+ (r+x_n) < m+1$
$$(1+x_1)\dots(1+x_{n-1})(1+x_n) \geq (1+x_n)(1+r)2^m \geq (1+r+x_n)2^m$$
which follows from the simple case for $n=2$ above.

2. The overflowing case, if $1 < x_n + r = 1 + r_2$ then $x_1 + \dots + x_n = (m+1) + r_2$

$$(1+x_1) \dots (1+x_{n-1})(1+x_n) \geq (1+x_n)(1+r)2^m$$
$$ \geq (1+r_2)2^{m+1}$$
which is similar to the overflowing case from $n=2$ above.

The equality cases can be determined as follows:
If all $x_i$s are zeros or ones, then the equality follows trivially. We assert then at most one of the $x_i$s are strictly between zero and one. Because if there are two of them, then both the simple case and the overflowing case in $n=2$ will never reach equality. The inductive step in the proof also depends on these two cases.

Integral Inequality

Prove that:

$$\left( \int_\pi^\infty\frac{\cos x}{x}\ dx\right)^{2} < \frac{1}{{\pi}^{2}} $$

Solution

Integrate by parts:

$$\int \frac{\cos x}{x} dx = \frac{\sin x}{x} + \int \frac{\sin x}{x^2} dx$$

So
$$\int_\pi^\infty\frac{\cos x}{x}\ dx = \int_\pi^\infty \frac{\sin x}{x^2} dx$$

And
$$| \int_\pi^\infty\frac{\cos x}{x}\ dx| = |\int_\pi^\infty \frac{\sin x}{x^2} dx|$$
$$\leq \int_\pi^\infty \frac{| \sin x |}{x^2} dx$$
$$\leq \int_\pi^\infty \frac{1}{x^2} dx $$
$$= \frac{1}{\pi}$$

Triangle Inequality

Given a triangle $ABC$ and a point $M$ inside the triangle.

Let $\alpha = \angle BMC, \beta = \angle AMC, \gamma = \angle AMB$

Prove that:
$$\frac{AM}{BM.CM} + \frac{BM}{CM.AM} + \frac{CM}{AM.BM} \geq -2 \left( \frac{\cos \alpha}{AM} + \frac{\cos \beta}{BM} + \frac{\cos \gamma}{CM} \right)$$

Solution In Progress

Let
$a = \frac{AM}{\sin \alpha}, b = \frac{BM}{\sin \beta}, c = \frac{CM}{\sin \gamma}$

Because $M$ is in the interior of the triangle, then $0 < \alpha, \beta, \gamma < \pi$ and thus $0 < \sin \alpha, \sin \beta, \sin \gamma \leq 1$. Thus $a,b,c > 0$. Without loss of generality, we may assume that $a \geq b \geq c$.

So we have:
$AM = a \sin \alpha, BM = b \sin \beta, CM = c \sin \gamma$

Substitute it to our inequality, and use the following shorthand:

$C_\alpha = \cos \alpha \sin \beta \sin \gamma$
$C_\beta = \sin \alpha \cos \beta \sin \gamma$
$C_\gamma = \sin \alpha \sin \beta \cos \gamma$

So our inequality becomes
$$ \iff a^2\sin^2 \alpha + b^2 \sin^2 \beta + c^2 \sin^2 \gamma +2 ( bc C_\alpha + ac C_\beta + ab C_\gamma ) \geq 0$$

Note the following identities:
$$C_\beta + C_\gamma = \sin \alpha \cos \beta \sin \gamma + \sin \alpha \sin \beta \cos \gamma = \sin \alpha \sin (\beta + \gamma) = \sin \alpha \sin (2 \pi - (\beta + \gamma)) = - \sin^2 \alpha$$

Similarly,
$$C_\alpha + C_\gamma = - \sin^2 \beta$$
$$C_\alpha + C_\beta = - \sin^2 \gamma$$

So that
$$C_\alpha = (\sin^2 \alpha - \sin^2 \beta - \sin^2 \gamma)/2$$
$$C_\beta = (\sin^2 \beta - \sin^2 \alpha - \sin^2 \gamma)/2$$
$$C_\gamma = (\sin^2 \gamma - \sin^2 \beta - \sin^2 \alpha)/2$$

Substituting back to our inequalities, we have:
$$\iff (a-b)(a-c)\sin^2 \alpha + (b-a)(b-c) \sin^2 \beta + (c-a)(c-b) \sin^2 \gamma \geq 0$$

It's also equivalent to:
$$\iff (a-b)^2 C_\gamma + (a-c)^2 C_\beta + (b-c)^2 C_\alpha \leq 0$$

Triangle Inequality

In a triangle $ABC$, let $a = BC, b = AC, c = AB$. For any point $M$ and real numbers $x,y,z$, show that

$(x+y+z)(xMA^2 + yMB^2 + zMC^2) \geq xyc^2 + xzb^2 + yza^2$

Solution
We shall show that the inequality above is equivalent to
$$(x \vec{MA} + y \vec{MB} + z \vec{MC})^2 \geq 0$$

Indeed, since:
$$2\vec{MA} \vec{MB} = MA^2 + MB^2 - c^2$$
so
$$2xy\vec{MA} \vec{MB} = xyMA^2 + xyMB^2 - xyc^2$$
$$2yz\vec{MB} \vec{MC} = yzMB^2 + yzMC^2 - yza^2$$
$$2zx\vec{MC} \vec{MA} = zxMC^2 + zxMA^2 - zxb^2$$

Adding them, we obtain:
$$RHS = \sum x(y+z)MA^2 - 2\sum xy \vec{MA} \vec{MB}$$

So
$$LHS - RHS = \sum x^2MA^2 + \sum x(y+z)MA^2 - RHS = \sum x^2 MA^2 + 2\sum xy \vec{MA} \vec{MB}$$
$$= (x \vec{MA} + y \vec{MB} + z \vec{MC})^2 \geq 0 $$

Inequality

For positive numbers $a,b,c>0$ such that $a+b+c=3$, find the maximum value of

$\frac{1}{2+a^2+b^2} + \frac{1}{2+a^2+c^2} + \frac{1}{2+b^2+c^2} $

Solution: Inequality

Original problem: http://dharmath.thehendrata.com/2009/12/02/inequality/

For positive real number $a,b,c$ prove that

$2\sqrt{ab+bc+ca} \leq \sqrt{3} \sqrt[3]{(a+b)(b+c)(c+a)}$

Inequality

For positive real number $a,b,c$ prove that

$2\sqrt{ab+bc+ca} \leq \sqrt{3} \sqrt[3]{(a+b)(b+c)(c+a)}$

solution: http://dharmath.thehendrata.com/2009/12/02/solution-inequality/

Solution: 3 Variable Inequality

Original problem: http://dharmath.thehendrata.com/2009/11/24/3-variable-inequality/

For $a,b,c > 0$, prove that:

$! (ab(a+b) + bc(b+c) + ca(c+a))^2 \geq 4abc(a+b+c)(a^2+b^2+c^2)$