Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Prove it

Proposition Let $P$ be a point inside $\triangle A B C$ , and let $P A=x, P B=y, P C=z$ , and the area of $\triangle A B C$ be $S$ . Then
$x+y+z \geqslant 2 \sqrt[4]{3 S^{2}} \text {. }$

where equality holds if and only if $a=b=c$ , and $P$ is the Fermat point.

Solution

Proof: We discuss in two cases.
(i) When $\max \{A, B, C\}$
$2 \sqrt{2 S}$
$=2 \sqrt{2} \sqrt{S}>2 \sqrt[4]{3} \sqrt{S}=4 \sqrt[4]{3 S^{2}},$

i.e., $x+y+z>2 \sqrt[4]{3 S^{2}}$ .
In summary: We can get $x+y+z \geqslant 2 \sqrt[4]{3 S^{2}}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.