Let $a$, $b$, $c$, $u$, $v$, $w$, and $p$ all be positive real numbers, and $$a^{1/p}+c^{1/p}\leq b^{1/p},\quad u^{1/(p+1)}+w^{1/(p+1)}\geq v^{1/(p+1)}$$ Prove that $ubc-vca+wab \ge 0$.
My answer: $$acv\leq ac(u^{\frac{1}{p+1}}+w^{\frac{1}{p+1}})^{p+1}=[(uc)^{\frac{1}{p+1}}a^{\frac{1}{p+1}}+(wa)^{\frac{1}{p+1}}c^{\frac{1}{p+1}}]^{p+1}$$
By Hölder's inequality, $$[(uc)^{\frac{1}{p+1}}a^{\frac{1}{p+1}}+(wa)^{\frac{1}{p+1}}c^{\frac{1}{p+1}}]^{p+1}\leq (uc+wa)(a^{\frac{1}{p}}+c^{\frac{1}{p}})\leq (uc+wa)b.$$ Done!
It is easy to see that this problem is related to Schur's inequality. I am finding its source (maybe from Olympiad) and its other answers.