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Find the smallest constant $K$satisfying the inequality $$x^{1\over 3}+y^{1\over 3} \le K(x+y)^{1\over 3}$$ The official proof makes the substitution $a=x^{1\over 3}$ and $b=y^{1\over 3}$, which does make the question a lot simpler, but then it just states that $K=4^{1\over 3}$ and proves this is true. I was wondering if anyone knew how one could determine this number instead of just knowing it and then proving it, thanks in advance!

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If you set $x=y$ and turn the inequality to an equation, the given equation in $K,x$ has the solution $K=4^{1/3}$, for all $x$.

Then this is a good candidate for a tentative proof.

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    $\begingroup$ Oh okay, so the smallest value of K occurs when both sides of the inequality are equal, that actually makes a lot more sense, thank you :) $\endgroup$ – Anthony Muleta Aug 16 '14 at 11:24
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I guess you assume $x,y\geq 0$, for otherwise it will not hold.

It is obvious that $f(x)=x^{1/3}$ is a concave function. Jensen's inequality shows that: $$ x^{1/3}/2+y^{1/3}/2\leq(x/2+y/2)^{1/3} $$ or rather $$ x^{1/3}+y^{1/3}\leq 2^{2/3}(x+y)^{1/3} $$ When $x,y\leq 0$, $f(x)=x^{1/3}$ becomes a convex function and we should rather have: $$ x^{1/3}+y^{1/3}\geq 2^{2/3}(x+y)^{1/3} $$

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