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

\(3p^2q+3p^2r+2q^3+2r^3 \leq 2\).

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

\(3p^2q+3p^2r+2q^3+2r^3 \leq 2\).

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## Comments

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TopNewestWe 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 \[3p^2q+3p^2r+2q^3+2r^3 = 3s-s^3 \leq 2\] – Kishlaya Jaiswal · 1 year, 5 months ago

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– Vishnu C · 1 year, 5 months ago

What inequality is \(3s-s^3 \leq 2\)?Log in to reply

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 · 1 year, 5 months ago

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– Vishnu C · 1 year, 5 months ago

Doh!Log in to reply

– Kishlaya Jaiswal · 1 year, 5 months ago

Thanks!Log in to reply

– Vishnu C · 1 year, 5 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!!!Log in to reply

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 · 1 year, 5 months ago

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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 · 1 year, 5 months ago

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