# Find maximum value by using AM-GM inequality

I have a problem: Find the maximum value of $$P=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}$$ with $$x,y,z>0$$. Is there anyway to solve this problem by using the AM-GM inequality ? Thank for your answer.

The first way, I tried to use AM-GM twice at the two sum in the denominator, but I get the sum of the fractions with square roots of $$xy, yz, zx$$ at its denominators.

The other way, I tried to factor the denominator and use both AM-GM and the Schwarz inequality but still get the same with the first way, with higher order

• what have you tried so far? Feb 26, 2019 at 6:21
• I tried to use AM-GM for the denominator but not sucess :| Feb 26, 2019 at 6:25
• @Hoàng Show please, how exactly you tried to make it. Actually, I solved your problem. If you want to see my solution, show us your trying. Feb 26, 2019 at 6:28
• The first way, i tried to use AM-GM twice at the two sum in the denominator, but i get the sum of the fractions with square roots of xy, yz, zx at its denominators. Feb 26, 2019 at 6:49
• The other way, i tried to factor the denominator and use both AM-GM and the Schwarz inequality but still get the same with the first way, with higher order Feb 26, 2019 at 6:51

For $$x=y=z=1$$ we get $$P=\frac{3}{16}.$$

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

Indeed, by AM-GM $$\sum_{cyc}\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}\leq\sum_{cyc}\frac{x^3y^4z^3}{(x^4+y^4)(2\sqrt{xyz^2})^3}=\sum_{cyc}\frac{\sqrt{x^3y^5}}{8(x^4+y^4)}.$$ Let $$\sqrt{\frac{x}{y}}=a$$, $$\sqrt{\frac{y}{z}}=b$$ and $$\sqrt{\frac{z}{x}}=c$$.

Thus, $$abc=1$$ and we need to prove that $$\sum_{cyc}\frac{a^3}{a^8+1}\leq\frac{3}{2}$$ or $$\sum_{cyc}\left(\frac{1}{2}-\frac{a^3}{a^8+1}\right)\geq0$$ or $$\sum_{cyc}\left(\frac{1}{2}-\frac{a^3}{a^8+1}-\frac{1}{2}\ln{a}\right)\geq0.$$ Now, let $$f(a)=\frac{1}{2}-\frac{a^3}{a^8+1}-\frac{1}{2}\ln{a}.$$

Thus, $$f'(a)=-\tfrac{(a-1)(a^{11}(a^4+a^3+a^2+a+1)-9a^8(a^2+a+1)-7a^3(a^4+a^3+a^2+a+1)-a^2-a-1)}{2a(a^8+1)^2}.$$ Since by the Descartes' rule of signs the polynomial $$a^{11}(a^4+a^3+a^2+a+1)-9a^8(a^2+a+1)-7a^3(a^4+a^3+a^2+a+1)-a^2-a-1$$ has an unique positive root (this root is $$a_1=1.56...$$) and $$f(a_1)>0$$,

we see that in $$a_1$$ the function $$f$$ has a local maximum,

which gives that$$f$$ decreases on $$[a_1,+\infty)$$ and there is an unique $$a_0>a_1$$, for which $$f(a_0)=0$$.

Easy to see that $$a_0=2.679...$$ and since $$a_{min}=1$$ and $$f(1)=0$$,

our inequality is proven for $$\max\{a,b,c\}\leq2.5$$

Let $$a\geq2.5$$.

Also, by AM-GM $$\frac{x^3}{x^8+1}=\frac{x^3}{3\cdot\frac{x^8}{3}+5\cdot\frac{1}{5}}\leq\frac{x^3}{8\sqrt[8]{\left(\frac{x^8}{3}\right)^3\left(\frac{1}{5}\right)^5}}=\frac{1}{8}\sqrt[8]{3^35^5}.$$ Id est, $$\sum_{cyc}\frac{a^3}{a^8+1}\leq\frac{1}{4}\sqrt[8]{3^35^5}+\frac{2.5^3}{2.5^8+1}=1.0423...<\frac{3}{2}$$ and we are done!

Another way based on Michael Rozenberg's solution.

If you prove $$\sum_{cyc}\frac{a^3}{a^8+1}\leq\frac{3}{2}$$ you can prove the lemma $$\frac{1}{a^2+a+1}+\frac{1}{b^2+b+1}+\frac{1}{c^2+c+1}\ge 1$$ for $$abc=1$$ and $$a;b;c\in R^+$$ $$(\text{Prove by C-S and} (a=xy/z^2;b=yz/x^2;c=xz/y^2))$$

Note that we have: $$\frac{a^3}{a^8+1}\le \frac{3\left(a^2+1\right)}{4\left(a^4+a^2+1\right)}$$

Or $$-\frac{\left(a-1\right)^2\left(3a^8+6a^7+16a^6+14a^5+16a^4+14a^3+12a^2+6a+3\right)}{4\left(a^2-a+1\right)\left(a^2+a+1\right)\left(a^8+1\right)}\le 0$$

It is simple now. :)