Minimize $(x+y)(y+z)(z+x)$ given $xyz(x+y+z) = 1$

Question

$x,y,z$ are positive reals and I am given $xyz(x+y+z) = 1$. Need to minimize $(x+y)(y+z)(z+x)$. Here is my approach.

Using AM-GM inequality

$$ (x+y) \geqslant 2 \sqrt{xy} $$ $$ (y+z) \geqslant 2 \sqrt{yz} $$ $$ (z+x) \geqslant 2 \sqrt{zx} $$

So, we have

$$ (x+y)(y+z)(z+x) \geqslant 8xyz $$

Also, I got

$$ \frac{x+y+z+(x+y+z)}{4} \geqslant \bigg[ xyz(x+y+z) \bigg] ^{1/4} $$

$$ \therefore x+y+z \geqslant 2 $$

But, I am stuck here. Any hints ?

Answer 1

Since $x+y+z \geqslant 3 \sqrt[3]{xyz}$, we have $xyz (x+y+z)\geqslant 3 (xyz)^{4/3}$.

Using the given condition, we have $1 \geqslant 3 (xyz)^{4/3}$. This is $xyz \leqslant \frac{1}{3^{3/4}} $

Also, we have $(x+y)(y+z)(z+x) = (x+y+z)(xy+ yz + zx) - xyz $

Now $ -xyz \geqslant - \frac{1}{3^{3/4}} $ and

$$ (xy+ yz + zx) \geqslant 3 (xyz)^{2/3} $$ $$ (x+y+z)(xy+ yz + zx) \geqslant 3 \frac{(x+y+z)xyz}{(xyz)^{1/3}} $$ $$ (x+y+z)(xy+ yz + zx) \geqslant \frac{3}{(xyz)^{1/3}} $$

But, we have

$$ xyz \leqslant \frac{1}{3^{3/4}} $$ $$ \therefore \frac{1}{xyz} \geqslant 3^{3/4} $$ $$ \therefore \frac{1}{(xyz)^{1/3}} \geqslant 3^{1/4} $$

So, we get

$$ (x+y+z)(xy+ yz + zx) \geqslant 3^{5/4} $$

So, we have

$$(x+y+z)(xy+ yz + zx) -xyz \geqslant 3^{5/4} - \frac{1}{3^{3/4}} $$

It follows that

$$ (x+y)(y+z)(z+x) \geqslant \frac{8}{3^{3/4}} $$

It can be seen that equality is achieved when $x=y=z$ in $xyz(x+y+z)=1$

Answer 2

$(x+y)(y+z)(z+x)=(z+x)(y(x+y+z)+xz)=(\frac{1}{zx}+zx)(x+z)$

now we can use $$\frac{1}{zx}+zx\ge 4{(\frac{1}{27{(xz)}^2})}^{1/4}$$

(HINT:$\frac{1}{zx}=\frac{1}{3zx}+\frac{1}{3zx}+\frac{1}{3zx}$)

also we can use $$x+z\ge 2\sqrt{xz}$$

Multiplying we get $$(x+y)(y+z)(z+x)\ge \frac{8}{3^{3/4}}$$

Answer 3

Setting $t=xyz(x+y+z),$ then $xy+yz+zx \geqslant \sqrt{3t}=\sqrt 3.$ Using known inequality $$(x+y)(y+z)(z+x) \geqslant \frac{8}{9}(x+y+z)(xy+yz+zx).$$ We have $$(x+y)(y+z)(z+x) \geqslant \frac{8}{9}(x+y+z)(xy+yz+zx)$$ $$\geqslant \frac{8}{9} \cdot \sqrt{3(xy+yz+zx)} \cdot (xy+yz+zx) = \frac{8}{9}\sqrt{3(xy+yz+zx)^3}$$ $$\geqslant \frac{8}{9}\sqrt{3\left(\sqrt{3}\right)^3} = \frac{8\sqrt[4]{3}}{3}.$$ Equality occur when $x=y=z=\frac{1}{\sqrt[4]{3}}.$

Answer 4

For $x=y=z=\frac{1}{\sqrt[4]3}$ we get a value $\frac{8}{\sqrt[4]{27}}.$

We'll prove that it's a minimal value.

Indeed, we need to prove that $$\prod_{cyc}(x+y)\geq\frac{8}{\sqrt[4]{27}}$$ or $$27\prod_{cyc}(x+y)^4\geq4096x^3y^3z^3(x+y+z)^3.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $$(9uv^2-w^3)^4\geq4096u^3w^9$$ or $f(w^3)\geq0,$ where $$f(w^3)=(9uv^2-w^3)^4-4096u^3w^9.$$ But it's obvious that $f$ decreases, which says that it's enough to prove our inequality for a maximal value of $w^3$, which by $uvw$ happens for equality case of two variables.

Since the last inequality is symmetric and homogeneous, it's enough to assume $y=z=1$ and we need to prove that: $$27(x+1)^8\geq256x^3(x+2)^3,$$ which is true by AM-GM: $$27(x+1)^8=27(x^2+2x+1)^4=27\left(3\cdot\frac{x^2+2x}{3}+1\right)^4\geq$$ $$\geq27\left(4\sqrt[4]{\left(\frac{x^2+2x}{3}\right)^3\cdot1}\right)^4=256x^3(x+2)^3$$ and we are done!