Find the minimum value of $P=\sum _{cyc}\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}$

Question

For $x>0$, $y>0$, $z>0$ and $x+y+z=3$ find the minimize value of $$P=\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}+\frac{\left(y+1\right)^2\left(z+1\right)^2}{x^2+1}+\frac{\left(z+1\right)^2\left(x+1\right)^2}{y^2+1}$$

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We have: $P=\left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{1}{\left(z+1\right)^2\left(z^2+1\right)}+\frac{1}{\left(y+1\right)^2\left(y^2+1\right)}+\frac{1}{\left(x+1\right)^2\left(x^2+1\right)}\right)$

$\ge \left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{1}{2\left(z^2+1\right)^2}+\frac{1}{2\left(x^2+1\right)^2}+\frac{1}{2\left(y^2+1\right)^2}\right)$

$\ge \left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{9}{2\left(\left(z^2+1\right)^2+\left(y^2+1\right)^2+\left(x^2+1\right)^2\right)}\right)$

I can't continue. Help

Answer 1

Let $x=y=z=1$.

Hence, $P=24$.

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

Indeed, by C-S $$\sum_{cyc}\frac{(x+1)^2(y+1)^2}{z^2+1}=\sum_{cyc}\frac{(x+1)^2(y+1)^2(x+y)^2}{(z^2+1)(x+y)^2}\geq\frac{\left(\sum\limits_{cyc}(x+1)(y+1)(x+y)\right)^2}{\sum\limits_{cyc}(z^2+1)(x+y)^2}.$$ Thus, it remains to prove that $$\left(\sum\limits_{cyc}(x+1)(y+1)(x+y)\right)^2\geq24\sum\limits_{cyc}(z^2+1)(x+y)^2.$$

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, $u=1$ and$$\sum\limits_{cyc}(x+1)(y+1)(x+y)=\sum_{cyc}(x^2y+x^2z+2x^2+2xy+2x)=$$ $$=9uv^2-3w^3+2u(9u^2-6v^2)+6uv^2+6u^3=3(8u^3+uv^2-w^3);$$ $$\sum\limits_{cyc}(z^2+1)(x+y)^2=2\sum_{cyc}(x^2y^2+x^2yz+x^2u^2+xyu^2)=$$ $$=2(9v^4-6uw^3+3uw^3+9u^4-6u^2v^2+3u^2v^2)=6(3u^4-u^2v^2+3v^4-uw^3).$$ Id est, it's enough to prove that $f(w^3)\geq0$, where $$f(w^3)=(8u^3+uv^2-w^3)^2-16(3u^6-u^4v^2+3u^2v^4-u^3w^3).$$ Now, $$f'(w^3)=-2(8u^3+uv^2-w^3)+16u^3=2w^3-2uv^2\leq0,$$ which says that $f$ is decreasing function.

Thus, it's enough to prove our inequality for a maximal value of $w^3$,

which happens for equality case of two variables.

Let $y=x$ and $z=3-2x$.

Hence, $$\left(\sum\limits_{cyc}(x+1)(y+1)(x+y)\right)^2\geq24\sum\limits_{cyc}(z^2+1)(x+y)^2$$ gives $$(x-1)^2(x^4-2x^3-11x^2+24x+4)\geq0.$$ which is obvious.

Done!

Answer 2

The following quantity is cleary positive \begin{eqnarray*} 4.5\sum_{cyc}(x-y)^8+146\sum_{cyc}xy(x-y)^6+1122\sum_{cyc}x^2y^2(x-y)^4+3678\sum_{cyc}x^3y^3(x-y)^2 \\ +584\sum_{cyc}xyz^2(xy-z^2)^2+2836\sum_{cyc}x^4(z-y)^4+786\sum_{perm}xy^2z^3(x-z)^2 \\ +12038\sum_{cyc}x^4yz(z-y)^2+4288\sum_{cyc}x^2y^2z^2((z-y)^2 \end{eqnarray*} Setting it to be greater than zero and rearranging gives \begin{eqnarray*} (4x+y+z)^2(x+4y+z)^2(9x^2+(x+y+z)^2)(9y^2+(x+y+z)^2)+ \\ (x+y+4z)^2(x+4y+z)^2(9z^2+(x+y+z)^2)(9y^2+(x+y+z)^2)+ \\ (4x+y+z)^2(x+y+4z)^2(9x^2+(x+y+z)^2)(9z^2+(x+y+z)^2)\\ \geq 24(9x^2+(x+y+z)^2)(9z^2+(x+y+z)^2)(9y^2+(x+y+z)^2)(x+y+z)^2) \\ \end{eqnarray*} This can rearranged to \begin{eqnarray*} \sum_{cyc}(3x+\color{red}{x+y+z})^2(3y+\color{red}{x+y+z})^2(9x^2+(\color{red}{x+y+z})^2)(9y^2+(\color{red}{x+y+z})^2+ \\ \geq 24(9x^2+(\color{red}{x+y+z})^2)(9z^2+(\color{red}{x+y+z})^2)(9y^2+(\color{red}{x+y+z})^2)(\color{red}{x+y+z})^2 \\ \end{eqnarray*} Now substitute $x+y+z=3$ and divide by $(x^2+1)(y^2+1)(z^2+1)$ \begin{eqnarray*} \sum_{cyc} \frac{(x+1)^2(y+1)^2}{(z^2+1)} \geq 24 \\ \end{eqnarray*} So the maximium is $\color{red}{24}$.

The above algebra is justified by the following computer algebra that will work in reduce.

((4*x+1*y+1*z)^2*(1*x+4*y+1*z)^2*(9*x^2+(x+y+z)^2)*(9*y^2+(x+y+z)^2)+ (1*x+1*y+4*z)^2*(1*x+4*y+1*z)^2*(9*z^2+(x+y+z)^2)*(9*y^2+(x+y+z)^2)+ (4*x+1*y+1*z)^2*(1*x+1*y+4*z)^2*(9*x^2+(x+y+z)^2)*(9*z^2+(x+y+z)^2)-

24*(9*x^2+(x+y+z)^2)*(9*z^2+(x+y+z)^2)*(9*y^2+(x+y+z)^2)*(x+y+z)^2)/9-

(4.5*((x-y)^8+(y-z)^8+(z-x)^8)+

146*(xy(x-y)^6+yz(y-z)^6+zx(z-x)^6)+

1122*(x^2*y^2*(x-y)^4+y^2*z^2*(y-z)^4+z^2*x^2*(z-x)^4)+

3678*(x^3*y^3*(x-y)^2+y^3*z^3*(y-z)^2+z^3*x^3*(z-x)^2)+

584*xyz*(z*(xy-z^2)^2+x(yz-x^2)^2+y(z*x-y^2)^2)+

2836*(x^4*(z-y)^4+y^4*(x-z)^4+z^4*(y-x)^4)+

786*(x^1*y^2*z^3*(x-z)^2+x^1*y^3*z^2*(x-y)^2+x^2*y^1*z^3*(y-z)^2+x^2*y^3*z^1*(y-z)^2+x^3*y^2*z^1*(x-z)^2+x^3*y^1*z^2*(x-y)^2)+

12038*(x^4*yz(z-y)^2+y^4*xz(x-z)^2+z^4*yx(y-x)^2)+

4288*x^2*y^2*z^2*((z-y)^2+(x-z)^2+(y-x)^2) );