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For $\left(x+y\right)\left(x+z\right)\left(y+z\right)=144$, minimize $$P=5\left(x^2+y^2\right)+2z^2$$

I have no idea. Can you make a few suggestions?

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    where does $z$ appear in the cost function? – 2017-02-22
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    I'm pretty sure you mean $5(x^2+y^2)+2z^2$. – 2017-02-22
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    ok , thank for advice – 2017-02-22
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    Perhaps, $$P=(2x-y)^2+(2y+x)^2+z^2$$Can we use it though? – 2017-02-22
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    i think $x=y=2$ and $z=4$ – 2017-02-22

3 Answers 3

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Let $x=y=2$ and $z=4$.

Hence, $P=72$. We'll prove that it's a minimal value.

Indeed, we need to prove that $$5(x^2+y^2)+2z^2\geq72\left(\sqrt[3]{\frac{(x+y)(x+z)(y+z)}{144}}\right)^2.$$

It's enough to prove last inequality for non-negative variables.

Let $x+y=tz$.

Since $x^2+y^2\geq\frac{1}{2}(x+y)^2$ and $(x+z)(y+z)\leq\frac{1}{2}(x+y+2z)^2$, it remains to prove that $$\frac{5}{2}t^2+2\geq72\left(\sqrt[3]{\frac{\frac{t(t+2)^2}{4}}{144}}\right)^2$$ or $$(5t^2+4)^3\geq9t^2(t+2)^4,$$ which is C-S and AM-GM: $$(5t^2+4)^3=\left(\frac{(5+4)(5t^2+4)}{9}\right)^3\geq\left(\frac{(5t+4)^2}{9}\right)^3=$$ $$=\left(\frac{(3t+2(t+2))^2}{9}\right)^3\geq\left(\frac{\left(3\sqrt[3]{3t\cdot(t+2)^2}\right)^2}{9}\right)^3=9t^2(t+2)^4.$$ Done!

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HINT: prove that $$P=5(x^2+y^2)+2z^2\geq 72$$ and the equal sign will atained at $$(x,y,z)=(2,2,4)$$

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    what is the method you use to prove it – 2017-02-22
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    i used the Lagrange multiplier method – 2017-02-22
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    it's good idea tks – 2017-02-22
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By AM-GM

$1728=(3x+3y)(2x+2z)(2y+2z)\leq \left(\frac{5x+5y+4z}{3}\right)^3$

$\Rightarrow 5x+5y+4z\geq 36$

By Cauchy-Schwarz we have:

$18P=(5x^2+5y^2+2z^2)(5+5+8)\geq (5x+5y+4z)^2\geq 36^2$

$$\Rightarrow P\geq 72$$

Hence $P_{\min}=72\Leftrightarrow (x,y,z)=(2,2,4)$