Most of CMI's questions are proof questions. Here's one more:

If p, q, r are real numbers such that $p^2+q^2+r^2=1$ , then prove that

$3p^2q+3p^2r+2q^3+2r^3 \leq 2$ .

#Algebra #Inequalities

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Comments

We simplify the given expression as follows -

$\begin{aligned} 3p^2q+3p^2r+2q^3+2r^3 &=& 3p^2(q+r)+2(q+r)(q^2-qr+r^2) \\ &=& (q+r)(3p^2+2q^2+2r^2-2qr) \\ &=& (q+r)(3(p^2+q^2+r^2)-(q^2+r^2)-2qr) \\ &=& (q+r)(3-(q+r)^2) \\ &=& 3(q+r)-(q+r)^3 \\ \end{aligned}$

Now since, $q,r \in \mathbb{R}$ , then let $s=q+r$ and thus $s \in \mathbb{R}$ . We must obviously have $3p^2q+3p^2r+2q^3+2r^3 = 3s-s^3 \leq 2$

Kishlaya Jaiswal - 6 years ago

What inequality is $3s-s^3 \leq 2$ ?

vishnu c - 6 years ago

Consider $f(s) = 3s-s^3$ .

We have $f^\prime(s) = 3-3s^2 = 0 \Rightarrow s=\pm1$ . Observe that $f^{\prime\prime}(1) <0$ and so we should have maximum at $s=1$ . And hence $f(s)_{\text{max}}=2$

And therefore, $3s-s^3 \leq 2$

Kishlaya Jaiswal - 6 years ago

@Kishlaya Jaiswal – Doh!

vishnu c - 6 years ago

@Vishnu C – Thanks!

Kishlaya Jaiswal - 6 years ago

@Kishlaya Jaiswal – I'm going to look like a complete idiot for saying this, but this is the truth: I got all the way up to $(q+r)(3-(q+r)^2)$ in my simplification. And then I simply left the question. I didn't know how to proceed any further. DAMN IT!!!

vishnu c - 6 years ago

Very nice proof @Kishlaya Jaiswal . It boils down to recognizing that the expression is a function of $q + r$ , from which the result follows.

Calvin Lin Staff - 6 years ago

Thank You Sir.

Actually at first sight, I was thinking if we can manipulate the expression so as to form function in $p^2,q^2,r^2$ so that we can use the condition $p^2+q^2+r^2=1$ alongwith AM-GM Inequality. But that didn't worked. It was then I realized that it's a function in $q+r$

Kishlaya Jaiswal - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$