How find this minimum

Question

Help me!

Let $x,y,z\ge0$ such that: $xy+yz+zx=1$.

Find the minimum value of: $A=\dfrac{1}{x^2+y^2}+\dfrac{1}{y^2+z^2}+\dfrac{1}{z^2+x^2}+\dfrac{5}{2}(x+1)(y+1)(z+1)$

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I found minimum value of $A$ is $\dfrac{25}{2}$ iff $(x,y,z)=(1,1,0)$ or any permutation. But I can't solve that.

Answer 1

first to prove:

$\dfrac{1}{x^2+y^2}+\dfrac{1}{y^2+z^2}+\dfrac{1}{z^2+x^2} \ge \dfrac{5(1+xyz)}{2}$ ...(1)

$A \ge \dfrac{25}{2} \implies x+y+z+2xyz\ge 2 $...(2)

(2) is easy to prove. But (1) is hard. I have a ugly solution and I think to wait some days to post it if there is no better way.

Answer 2

For $z=0$ and $x=y=1$ we get a value $\frac{25}{2}$.

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

For which we'll prove firstly that: $$\sum_{cyc}\frac{1}{x^2+y^2}\geq\frac{5}{2}+5xyz.$$ Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v\geq0$, and $xyz=w^3$.

Hence, $3v^2=1$ and we need to prove that $$\frac{\sum\limits_{cyc}(x^4+3x^2y^2)}{\prod\limits_{cyc}(x^2+y^2)}\geq\frac{5}{2}+5xyz$$ or $f(u)\geq0$, where $$f(u)=\frac{3v^2((9u^2-6v^2)^2+2(9v^4-6uw^3))}{\prod\limits_{cyc}(9u^2-6v^2-a^2)}-\frac{5}{2}-\frac{5w^3}{3\sqrt3v^3}$$ or $$f(u)=\frac{9v^2(27u^4-36u^2v^2+18v^4-4uw^3)}{81u^2v^4-54v^6-54u^3w^3+36uv^2w^3-w^6}-\frac{5}{2}-\frac{5w^3}{3\sqrt3v^3}.$$

Now, we'll prove that $f$ is an increasing function. For which we need to prove that $$(108u^3-72uv^2-4w^3)(81u^2v^4-54v^6-54u^3w^3+36uv^2w^3-w^6)\geq$$ $$\geq(27u^4-36u^2v^2+18v^4-4uw^3)(162uv^4-162u^2w^3+36v^2w^3)$$ or $$2(27u^3-18uv^2-w^3)(81u^2v^4-54v^6-54u^3w^3+36uv^2w^3-w^6)\geq$$ $$\geq9(27u^4-36u^2v^2+18v^4-4uw^3)(9uv^4-9u^2w^3+2v^2w^3).$$ But $$2(27u^3-18uv^2-w^3)(81u^2v^4-54v^6-54u^3w^3+36uv^2w^3-w^6)-$$ $$-9(27u^4-36u^2v^2+18v^4-4uw^3)(9uv^4-9u^2w^3+2v^2w^3)=$$ $$=2187u^5v^4-2916u^3v^6+486uv^8-729u^6w^3+486u^4v^2w^3+$$ $$+972u^2v^4w^3-216v^6w^3-270u^3w^6+36uv^2w^6+2w^9\geq$$ $$\geq2187u^5v^4-2916u^3v^6+486uv^8-729u^6w^3+486u^4v^2w^3+972u^2v^4w^3-216v^6w^3-270u^3w^6=$$ $$=27(81u^5v^4-108u^3v^6+18uv^8-27u^6w^3+18u^4v^2w^3+36u^2v^4w^3-8v^6w^3-10u^3w^6)\geq$$ $$\geq27(81u^5v^4-108u^3v^6+18uv^8-27u^6w^3+18u^4v^2w^3+18u^2v^4w^3)=$$ $$=243u(9u^4v^4-12u^2v^6+2v^8-3u^5w^3+2u^3v^2w^3+2uv^4w^3).$$ Thus, it remains to prove that $$9u^4v^4-12u^2v^6+2v^8-3u^5w^3+2u^3v^2w^3+2uv^4w^3\geq0,$$ which is a linear inequality of $w^3$, which says that it's enough to prove the last inequality

for an extremal value of $w^3$, which happens in the following cases.

1. $w^3=0$.

Let $z=0$ and $y=1$.

We need to prove that $$\frac{(x+1)^4x^2}{81}-\frac{4(x+1)^2x^3}{81}+\frac{2x^4}{81}\geq0$$ or $$(x-1)^2(x+1)^2+2x^2\geq0,$$ which is obvious.

2. $y=z=1$.

In this case we obtain: $$(x-1)^2(3x^4+4x^3+2)\geq0.$$ Id est, $f$ is an increasing function, which says that it's enough to prove $f(u)\geq0$

for a minimal value of $u$, which happens for equality case of two variables.

Let $y=x$.

Hence, $z=\frac{1-x^2}{2x}$ and we need to prove that $$(x-1)^2(x+1)(25x^6+25x^5+15x^4-10x^3-5x^2+x+1)\geq0,$$ which is true because by AM-GM $25x^6-10x^3+1\geq0$ and $$25x^5+15x^4-5x^2+x=x\left(25x^4+15x^3+5\cdot\frac{1}{5}-5x\right)\geq$$ $$\geq x^2\left(7\sqrt[7]{\frac{25\cdot15}{5^5}}-5\right)\geq0.$$ Thus, it remains to prove that $$\frac{5}{2}+5xyz+\frac{5}{2}(x+1)(y+1)(z+1)\geq\frac{25}{2}$$ or $$3xyz+x+y+z\geq2,$$ which is a linear inequality of $u$.

Thus, again it's enough to prove the last inequality for $y=x$ and $z=\frac{1-x^2}{2x},$ where $0<x\leq1$, which gives $$(1-x)(3x^3+3x^2-3x+1)\geq0.$$ Done!