<s>[INST] Finding maximum with two constraints Let $a$,$b$,$c$ be positive real numbers satisfying $$a+b+c=1$$ $$a^2+b^2+c^2=\frac{3}{8}$$ Find the maximum value of $$a^3+b^3+c^3$$
Using the well known $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, the problem changes to finding maximum of $abc$ but in this case AM-GM, cauchy schwarz or holder is not easy. Can someone help me? (I don't want lagrange multiplier solution)
 [/INST] Let $a+b+c=3u,$ $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $u=\frac{1}{3}$, $$2(ab+ac+bc)=(a+b+c)^2-(a^2+b^2+c^2),$$ which gives
$$v^2=\frac{5}{48}.$$
Now, $$(a-b)^2(a-c)^2(b-c)^2\geq0$$ gives
$$3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6\geq0,$$ which gives
$$3uv^2-2u^3-2\sqrt{(u^2-v^2)^3}\leq w^3\leq3uv^2-2u^3+2\sqrt{(u^2-v^2)^3},$$
which says $$w^3\leq3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}=\frac{1}{32}.$$
Id est, $$a^3+b^3+c^3=27u^3-27uv^2+3w^3\leq27\cdot\frac{1}{27}-27\cdot\frac{1}{3}\cdot\frac{5}{48}+\frac{3}{32}=\frac{5}{32}.$$
The equality occurs for $a=b$, which gives $$(a,b,c)=\left(\frac{5}{12},\frac{5}{12},\frac{1}{6}\right),$$ which says that we got a maximal value.
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