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# 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

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

Note by Vishnu C
2 years, 10 months ago

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We simplify the given expression as follows -

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

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

- 2 years, 10 months ago

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

- 2 years, 10 months 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, $$3s-s^3 \leq 2$$

- 2 years, 10 months ago

Doh!

- 2 years, 10 months ago

Thanks!

- 2 years, 10 months ago

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!!!

- 2 years, 10 months 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.

Staff - 2 years, 10 months 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 $$q+r$$

- 2 years, 10 months ago