How to solve this step and are there any tips for this method?

Question

This is the problem: Let $x,y,z$ be real number, prove that: $(x+y)^4 +(y+z)^4 +(z+x)^4 -\frac{4}{7} (x^4+y^4+z^4) \geq 0$. (Vietnam TST 1996)

All the solutions I found are very complicated and not able to think about during the test, so I decide to solve it by myself.

Solution:

I put: $f(x;y;z)=(x+y)^4 +(y+z)^4 +(z+x)^4 -\frac{4}{7} (x^4+y^4+z^4)$

So: $f(\frac{x+y}{2};\frac{x+y}{2};z)=(a+b)^4+(c+\frac{a+b}{2})^4+(c+\frac{a+b}{2})^4-\frac{4}{7}((\frac{a+b}{2})^4+(\frac{a+b}{2})^4+c^4)$

In this method, I tried to make: $f(x;y;z) \geq f(\frac{x+y}{2};\frac{x+y}{2};z) \Leftrightarrow f(x;y;z) - f(\frac{x+y}{2};\frac{x+y}{2};z) \geq0$

And I put $x,y,z$ into this inequality and this is what I got :

$\Leftrightarrow \frac{1}{2}(x-y)^2[(x+z)^2+(x+z)(y+z)+(y+z)^2+z(x+y+2z)]+\frac{2}{7}(x-y)^2[-t(x+y)-t^2-(y^2+xy+x^2)] \geq 0$

where $t=\frac{x+y}{2}$

Can anyone help me with the next step with natural solution please? Thank you a lot!

Answer 1

Yes, MV helps.

If $x$, $y$ and $z$ are non-negatives, so our inequality is obvious.

Let there be negative variables.

Since our inequality is not changed after substitutions $x\rightarrow-x$, $y\rightarrow-y$ and $z\rightarrow-z$, it's enough to prove our inequality for $x\geq0$, $y\geq0$ and $z\leq0$.

Now, replace $z$ on $-z$ and we need to prove that $f(x,y,z)\geq0,$ where $$f(x,y,z)=(x-z)^4+(y-z)^4+(x+y)^4-\frac{4}{7}(x^4+y^4+z^4)$$ for non-negatives $x$, $y$ and $z$.

1. $z\geq\left(\frac{1}{2}+\frac{1}{\sqrt7}\right)(x+y).$

Thus, $$f(x,y,z)-f\left(\frac{x+y}{2},\frac{x+y}{2},z\right)=3(x-y)^2\left(z^2-(x+y)z+\frac{7x^2+10xy+7y^2}{56}\right)=$$ $$=3(x-y)^2\left(\left(z-\frac{x+y}{2}\right)^2-\frac{7x^2+18xy+7y^2}{56}\right)\geq$$ $$\geq3(x-y)^2\left(\frac{(x+y)^2}{7}-\frac{7x^2+18xy+7y^2}{56}\right)=\frac{3(x-y)^4}{56}\geq0,$$ which says that it's enough to prove our inequality for $x=y$, which gives a true inequality(see my previous solution).

2. $z\leq\left(\frac{1}{2}+\frac{1}{\sqrt7}\right)(x+y).$

In this cases it's enough to prove that $$(x-z)^4+(y-z)^4+(x+y)^4\geq\frac{4}{7}\left(x^4+y^4+\left(\frac{1}{2}+\frac{1}{\sqrt7}\right)^4(x+y)^4\right)$$ and since $$(x+y)^4\geq x^4+y^4,$$ it's enough to prove that $$3>4\left(\frac{1}{2}+\frac{1}{\sqrt7}\right)^4$$ or $$355>88\sqrt7,$$ which is true because $$355>88\sqrt9>88\sqrt7.$$

Answer 2

Your method (Mixing Variables method) is not so good here because we need to work with real numbers.

I think the best way here it's using $uvw$.

Indeed, this inequality is fourth degree, which says it's a linear inequality of $xyz$ and from here by $uvw$ it's enough to check $y=z=1$ (for $y=z=0$ our inequality is obviously true), which gives $$5x^4+28x^3+42x^2+28x+59\geq0,$$ which is true because $$5x^4+28x^3+42x^2+28x+59=5\left(x^2+\frac{14}{5}x-\frac{1}{5}\right)^2+\frac{6}{5}(2x+7)^2>0.$$ About $uvw$ see here: https://artofproblemsolving.com/community/c6h278791

Answer 3

Remark: Here is an alternative proof.

Since the inequality does not change if $(x, y, z)$ is substituted with $(-x, -y, -z)$, we assume that at most one of $x, y, z$ is negative.

If $x, y, z \ge 0$, clearly the inequality is true.

It remains to prove the case that $x \ge y \ge 0$ and $z < 0$ (due to symmetry).

Since the inequality is homogeneous, assume that $z = -1$. It suffices to prove that $$(x + y)^4 + (y - 1)^4 + (x - 1)^4 - \frac47(x^4 + y^4 + 1) \ge 0. \tag{1}$$

Case 1: $x \ge 1$

Using $(x + y)^4 - x^4 \ge (1 + y)^4 - 1^4$ (easy), we have \begin{align*} \mathrm{LHS}_{(1)} &= (x + y)^4 - x^4 + \frac37x^4 + (y - 1)^4 + (x - 1)^4 - \frac47(y^4 + 1)\\ &\ge (1 + y)^4 - 1^4 + \frac37 + (y - 1)^4 + (1 - 1)^4 - \frac47(y^4 + 1)\\ &= (1 + y)^4 + (y-1)^4 - \frac87 - \frac47y^4 \\ &\ge 2 + 2y^4 - \frac87 - \frac47y^2\\ & > 0 \end{align*} where we use $(1+y)^4 + (1 - y)^4 \ge 2 + 2y^4$ (easy).

Case 2: $x < 1$

We have \begin{align*} \mathrm{LHS}_{(1)} &\ge \frac{(x + y + 1 - y)^4}{8} + (x - 1)^4 - \frac47(x^4 + x^4 + 1) \\ &= \frac{(1+x)^4}{8} + (1-x)^4 - \frac87x^4 - \frac47\\ &\ge 0 \end{align*} where we use $a^4 + b^4 \ge (a + b)^4/8$ for all $a, b \ge 0$ (using the power mean inequality).

We are done.