How would you find the maximum value of $\frac{1}{(2a+b+c)^3}+\frac{1}{(2b+c+a)^3}+\frac{1}{(2c+a+b)^3}$ when $abc=1$ and $a,b,c$ are positive real numbers? (if it exists)
Again inspired by: The maximum of $\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}$.
If $a=b=c=1$ then we'll get a value $\frac{3}{64}.$
We'll prove that it's a maximal value.
Indeed, by C-S and AM-GM $$\sum_{cyc}\frac{1}{(2a+b+c)^3}=$$ $$=\sum_{cyc}\frac{1}{8a^3+b^3+c^3+9a(ab+ac+bc)+3bc(a+b+c)+3a(2(b^2+c^2)+a(b+c))}\leq$$ $$\leq\tfrac{1}{(10+27+9+18)^2}\sum_{cyc}\left(\tfrac{10^2}{8a^3+b^3+c^3}+\tfrac{27^2}{9a(ab+ac+bc)}+\tfrac{9^2}{3bc(a+b+c)}+\tfrac{18^2}{3a(2(b^2+c^2)+a(b+c))}\right)\leq$$ $$\leq\tfrac{1}{64^2}\sum_{cyc}\left(\tfrac{100}{8a^3+b^3+c^3}+\tfrac{81}{a(ab+ac+bc)}+\tfrac{27}{bc(a+b+c)}+\tfrac{108}{a((b+c)^2+a(b+c))}\right)=$$ $$=\frac{25}{1024}\sum_{cyc}\frac{1}{8a^3+b^3+c^3}+\frac{27}{1024abc}+\frac{27}{1024}\sum_{cyc}\frac{bc}{abc(b+c)(a+b+c)}\leq$$ $$\leq\frac{25}{1024}\sum_{cyc}\frac{1}{8a^3+b^3+c^3}+\frac{27}{1024abc}+\frac{27}{1024}\sum_{cyc}\frac{\frac{(b+c)^2}{4}}{abc(b+c)(a+b+c)}=$$ $$=\frac{25}{1024}\sum_{cyc}\frac{1}{8a^3+b^3+c^3}+\frac{81}{2048abc}.$$ Thus, it's enough to prove that $$\frac{25}{1024}\sum_{cyc}\frac{1}{8a^3+b^3+c^3}+\frac{81}{2048abc}\leq\frac{3}{64abc}$$ or $$\sum_{cyc}\frac{1}{8a^3+b^3+c^3}\leq\frac{3}{10abc}$$ or $$\sum_{cyc}\frac{1}{8a+b+c}\leq\frac{3}{10},$$ where $a$, $b$ and $c$ are positives such that $abc=1$.
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that $f(v^2)\geq0$, where $f$ is a linear function.
But the linear function gets a minimal value for an extremal value of $v^2$,
which happens for equality case of two variables.
Let $ b=a$ and $c=\frac{1}{a^2}$.
We obtain $$(a-1)^2(27a^4-11a^3-49a^2+24a+12)\geq0,$$ which is easy enough: $$27a^4-11a^3-49a^2+24a+12=$$ $$=11(a^4-a^3-a^2+a)+12(a^4-2a^2+1)+4a^4-14a^2+13a\geq$$ $$\geq a\left(4a^3+2\cdot\frac{13}{2}-14a\right)\geq a^2\left(3\sqrt[3]{13^2}-14\right)>0.$$ Done!
Also, we can use the following way.
Let $a^{\frac{3}{4}}=x$, $b^{\frac{3}{4}}=y$ and $c^{\frac{3}{4}}=z$.
Hence, $xyz=1$ and by AM-GM, P-M and AM-GM we obtain: $$\sum_{cyc}\frac{1}{(2a+b+c)^3}=\sum_{cyc}\frac{1}{a+3\cdot\frac{a+b+c}{3}}\leq\sum_{cyc}\frac{1}{64\left(\sqrt[4]{a\left(\frac{a+b+c}{3}\right)^3}\right)^3}\leq$$ $$\leq\sum_{cyc}\frac{1}{64x\left(\frac{x+y+z}{3}\right)^3}=\frac{xy+xz+yz}{64\left(\frac{x+y+z}{3}\right)^2\cdot\frac{x+y+z}{3}}\leq\frac{3}{64}.$$ Another way.
By AM-GM, AM-GM and AM-GM we obtain: $$\sum_{cyc}\frac{1}{(2a+b+c)^3}=\sum_{cyc}\frac{1}{(2a+b+c)^2(2a+b+c)}\leq\sum_{cyc}\frac{1}{16a\sqrt{bc}(2a+b+c)}=$$ $$=\sum_{cyc}\frac{\sqrt{bc}}{16(a+b+a+c)}\leq\frac{1}{32}\sum_{cyc}\sqrt{\frac{bc}{(a+b)(a+c)}}=$$ $$=\frac{1}{32}\sum_{cyc}\sqrt{\frac{b}{a+b}}\cdot\sqrt{\frac{c}{a+c}}\leq\frac{1}{64}\sum_{cyc}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)=$$ $$=\frac{1}{64}\sum_{cyc}\left(\frac{b}{a+b}+\frac{a}{b+a}\right)=\frac{3}{64}.$$
Let $3s=a+b+c\geqslant3$, as $abc=1$; then we need to show $$\sum_{cyc} \frac1{(3s+a)^3}\leqslant \frac3{16}$$
Let $f(x)=\dfrac1{16}-\dfrac1{(3s+x)^3}-\dfrac3{256s^3}\log x$, it is enough to show for $x\in (0,3s), s\geqslant 1$ that $f(x)\geqslant0$.
That is not hard as $f’(x)\leqslant 0$ and $f(3s)$ increases with $s$ with a positive value for $s=1$.
