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if $p^2+q^2+r^2=5$ and $p,q,r$ all are real number,

then maximum value of $(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$

what i try . Expanding $(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$

$41p^2+41q^2+25r^2-24pq-40qr-30pr$

$25\times 5+16p^2+16q^2-24pq-40qr-30pr$

How i use inequality to find maximum of given expression

Help me please

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    $\begingroup$ en.wikipedia.org/wiki/Lagrange_multiplier $\endgroup$ – Federico Dec 17 '18 at 19:38
  • $\begingroup$ The maximum is $250$, achieved at $p=-\frac{4}{\sqrt{5}},\ q =\frac{3}{\sqrt{5}},\ r=0$. $\endgroup$ – Federico Dec 17 '18 at 19:40
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    $\begingroup$ you expanded the thing incorrectly. For example, the coefficient of $p^2$ is actually $16+25=41$ $\endgroup$ – Will Jagy Dec 17 '18 at 20:01
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    $\begingroup$ Please check out our guide for new askers. This is very much missing in context. Are you expected to just use some trickery (high school level tools extended for contests), did this come up in a multivariable calculus course when we could expect you to be familiar with Lagrange multipliers? Et cetera. $\endgroup$ – Jyrki Lahtonen Dec 17 '18 at 20:35
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We'll prove that $250$ it's a maximal value.

Indeed, $$250\geq(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$ it's $$50(p^2+q^2+r^2)\geq(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$ or $$(3p+4q+5r)^2\geq0.$$ The equality occurs for $p^2+q^2+r^2=5$ and $3p+4q+5r=0.$

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