# maximum and minimum of $\frac{a^4+b^4+c^4}{(a+b+c)^4},$

Finding maximum and minimum of $\displaystyle \frac{a^4+b^4+c^4}{(a+b+c)^4},$ Where $a,b,c>0$ and $(a+b+c)^3=32abc$

$\bf{My Attempt}$ with AM-GM inequality $\displaystyle \frac{a+b+c}{3}\geq (abc)^{\frac{1}{3}}\Rightarrow (a+b+c)^3\geq 27abc$

Could some help me to solve it , thanks

Let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2$.
Hence, $t\geq1$ and the condition gives $$w^3=\frac{27u^3}{32}.$$ Also, $$(a-b)^2(a-c)^2(b-c)^2\geq0$$ gives $$3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6\geq0$$ or
$$3u^2v^4-4v^6-4u^3\cdot\frac{27u^3}{32}+6uv^2\cdot\frac{27u^3}{32}-\left(\frac{27u^3}{32}\right)^2\geq0$$ or $$-4185u^6+5184u^4v^2+3072u^2v^4-4096v^6\geq0$$ or $$4185t^3-5184t^2-3072t+4096\leq0$$ or $$(15t-16)(279t^2-48t-256)\leq0,$$ which with $t\geq1$ gives $$\frac{8+40\sqrt5}{93}\leq t\leq\frac{16}{15}.$$ Thus, $$f\left(\frac{1}{t}\right)=\frac{a^4+b^4+c^4}{(a+b+c)^4}=\frac{81u^4-108u^2v^2+18v^4+12uw^3}{81u^4}=$$ $$=\frac{81u^4-108u^2v^2+18v^4+12u\cdot\frac{27u^3}{32}}{81u^4}=\frac{81t^2-96t+16}{72t^2}=\frac{1}{72}\left(\frac{16}{t^2}-\frac{96}{t}+81\right)$$
and since $\left(\frac{1}{t}\right)_{min}=\frac{96}{2\cdot16}=3,$ we see that $f$ decreases on $\left[\frac{15}{16},\frac{93}{8+40\sqrt5}\right].$
Id est, $$\max\frac{a^4+b^4+c^4}{(a+b+c)^4}=f\left(\frac{15}{16}\right)=\frac{9}{128}$$ and $$\min\frac{a^4+b^4+c^4}{(a+b+c)^4}=f\left(\frac{93}{8+40\sqrt5}\right)=\frac{383-165\sqrt5}{256}.$$ Done!