Saturday, July 14, 2018

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.


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 $y > 0$: $$h(y) = g(y) + g(\frac{y}{t})$$

Wednesday, May 16, 2018

Shifting matrix

An $n \times n$ matrix is filled with numbers, where there are $k$ number ones and the rest zero. The operation that is allowed on the matrix is to shift a single column down by one (so that each number in that column moves down by one, and the bottom most number goes to the top), or to shift a single row to the left by one.

Determine all $k$ such that, no matter how the initial condition is, through a series of operations, we can make it so that each column and each row has an even sum.

Tuesday, May 15, 2018

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


$$2(x^7+y^7)(x^7+z^7) = x^{14} + (x^{14} + 2x^7(y^7+z^7) + 2y^7z^7)$$ 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: $$ \frac{x}{x^2 + myz} + \frac{y}{y^2 + mxz} + \frac{z}{z^2 + mxy} \geq \frac{9}{(m+1)(x+y+z)}$$


First we show that for $m \geq 8$ it's true. By Cauchy: $$\frac{x}{x^2 + myz} + \frac{y}{y^2 + mxz} + \frac{z}{z^2 + mxy} \geq \frac{(x+y+z)^2}{x^3 + y^3 + z^3 +}$$ 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 $m \geq 8$.

Friday, May 11, 2018

Inequality a b c

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


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.


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


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.

Wednesday, May 9, 2018

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: $$(3a+1)(3b+1)(3c+1) \geq 3 \sqrt{3}(\sqrt{a} + \sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})$$


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 $a=b=c = 1/3$.