Monday, September 21, 2009

Inequality of The Day

For $a,b,x,y$ real numbers, prove that:

$(3a^2+2ab+5b^2)(3x^2 + 2xy + 5y^2) \geq (3ax + ay + bx + 5by)^2$

Generalization: for $a,b,x,y$ real numbers, and $P,Q,R$ real numbers such that $Q^2 < PR$, prove that:

$(Pa^2+2Qab+Rb^2)(Px^2 + 2Qxy + Ry^2) \geq (Pax + Qay + Qbx + Rby)^2$

First Solution:

Upon expansion,the inequality simplifies to

$PRa^2y^2 + 2Q^2abxy + PRb^2x^2 \geq Q^2a^2y^2 + Q^2b^2y^2+ 2PRabxy$

$(PR-Q^2)(a^2y^2 + b^2x^2 - 2abxy) \geq 0$

$(PR-Q^2)(ay-bx)^2 \geq 0$

Second Solution:

If we replace $P,Q,R$ with $-P,-Q,-R$ respectively, the problem does not change. So without loss of generality, we can assume that $P \geq 0$.

Now, let $f$ be a function from $\mathbb{R}^4 \to \mathbb{R}$ defined by:

$f(a,b,x,y) = Pax + Q ay + Qbx + Rby$

Now the problem is equivalent to showing that $f(a,b,a,b) f(x,y,x,y) \geq f(a,b,x,y)^2$.

First, we show that $f(a,b,a,b) = Pa^2 + 2Qab + Rb^2 \geq 0$ for all $a,b$. That is true because $P \geq 0$ and the discriminant of the quadratic form $4Q^2-4PR < 0$. Furthermore, $f(a,b,a,b) = 0$ if and only if $a=b=0$.

Now, we notice that

$f(a\pm a', b \pm b', x, y) = f(a,b,x,y) \pm f(a',b',x,y)$

and similarly,

$f(a,b,x \pm x', y \pm y') = f(a,b,x,y) \pm f(a,b,x',y')$

And also,

$f(a,b,x,y) = f(x,y,a,b)$

$f(ka,kb,x,y) = f(a,b,kx,ky) = kf(a,b,x,y)$ for any $k$ real number.

Now, if either $f(a,b,a,b)$ or $f(x,y,x,y)$ is zero, the desired inequality holds trivially. So we let:

$a' = a/\sqrt{f(a,b,a,b)}, b' = b/\sqrt{f(a,b,a,b)}$

$x' = x/\sqrt{f(x,y,x,y)}, y' = y/\sqrt{f(x,y,xy)}$

And,

$f(a'-x',b'-y', a'-x', b'-y') \geq 0$

$f(a'-x',b'-y',a',b') - f(a'-x',b'-y',x',y') \geq 0$

$f(a',b',a',b') - f(x',y',a',b') - f(a',b',x',y') + f(x',y',x',y') \geq 0$

$f(a',b',a',b') + f(x',y',x',y') \geq 2f(a',b',x',y')$

But $f(a',b',a',b') = f(a,b,a,b) / f(a,b,a,b) = 1$, and similarly $f(x',y',x',y') = 1$.

That means $f(a',b',x',y') \leq 1$, which translates to:

$f \left(\frac{a}{\sqrt{f(a,b,a,b)}}, \frac{b}{\sqrt{f(a,b,a,b)}}, \frac{x}{\sqrt{f(x,y,x,y)}}, \frac{y}{\sqrt{f(x,y,x,y)}} \right) \leq 1$

$f(a,b,x,y) \leq \sqrt{f(a,b,a,b)} \sqrt{f(x,y,x,y)}$

$f(a,b,x,y)^2 \leq f(a,b,a,b) f(x,y,x,y)$

Remark: while this approach is significantly lengthier than direct expansion (see First Solution), this technique can be used for any function $f$ that satisfies the properties:

$f(a,b,a,b) \geq 0$ and $f(a,b,a,b) = 0 \iff a=b=0$

$f(a+ka',b+kb',x,y) = f(a,b,x,y) + kf(a',b',x,y)$

$f(a,b,x,y) = f(x,y,a,b)$.

Such function is usually called the "inner product" of vectors $(a,b)$ and $(x,y)$, and it can be generalized to $\mathbb{R}^n$

Third Solution:

Just like in the Second Solution, we can assume $P > 0$ without loss of generality.

Let $f(t) = Pt^2 + 2Qt + R$. Since $PR > Q^2$, that means $f$ can be expressed as a sum of two squares:

$f(t) = P(t+ C)^2 + D^2$ where: $C = Q/P, D^2 = R - Q^2/P$

Thus, $Pa^2 + 2Qab + Rb^2 = P(a+Cb)^2 + D^2b^2$ and $Px^2 + 2Qxy + Ry^2 = P(x+Cy)^2 + D^2y^2$

Solution by Cauchy:

$LHS = (P(a+Cb)^2 + D^2b^2)(P(x+Cy)^2 + D^2y^2)$

$\geq (P(a+Cb)(x+Cy) + D^2by)^2$

$= (Pax + PCbx + PCay + PC^2by + D^2by)^2 = (Pax + Qbx + Qay + Rby)^2$

Fourth Solution:

This is a close variant of the Second Solution, but for the sake of completeness, I will include everything here.

If we replace $P,Q,R$ with $-P,-Q,-R$ respectively, the problem does not change. So without loss of generality, we can assume that $P \geq 0$.

Now, let $f$ be a function from $\mathbb{R}^4 \to \mathbb{R}$ defined by:

$f(a,b,x,y) = Pax + Q ay + Qbx + Rby$

Now the problem is equivalent to showing that $f(a,b,a,b) f(x,y,x,y) \geq f(a,b,x,y)^2$.

First, we show that $f(a,b,a,b) = Pa^2 + 2Qab + Rb^2 \geq 0$ for all $a,b$. That is true because $P \geq 0$ and the discriminant of the quadratic form $4Q^2-4PR < 0$. Furthermore, $f(a,b,a,b) = 0$ if and only if $a=b=0$.

Now, we notice that

$f(a\pm a', b \pm b', x, y) = f(a,b,x,y) \pm f(a',b',x,y)$

and similarly,

$f(a,b,x \pm x', y \pm y') = f(a,b,x,y) \pm f(a,b,x',y')$

And also,

$f(a,b,x,y) = f(x,y,a,b)$

$f(ka,kb,x,y) = f(a,b,kx,ky) = kf(a,b,x,y)$ for any $k$ real number.

So,

$f(a+tx,b+ty,a+tx,b+ty) \geq 0$ for all $t,a,x,b,y$

$f(a+tx,b+ty,a,b) + tf(a+tx,b+ty,x,y) \geq 0$

$f(a,b,a,b)+tf(a,b,x,y) + tf(a,b,x,y) + t^2f(x,y,x,y) \geq 0$

So the polynomial $p(t) = f(x,y,x,y)t^2 + 2f(a,b,x,y)t + f(a,b,a,b)$ is never negative. That means $p(t)$ is a quadratic polynomial whose discriminant is less than zero.

$f(a,b,x,y)^2 \leq f(x,y,x,y)f(a,b,a,b)$