Track / Stage 5 / 337 of 400 #937 of 1964

Problem 937

AIME late

Algebra Difficulty 5.8 Prove it

Question 5 Let $a, b, c \in \mathbf{R}_{+}$ , and $a^{2}+b^{2}+c^{2}+a b c=4$ , prove: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant a+b+c$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Proof: From the common inequality $x y+y z+z x \leqslant \frac{1}{3}(x+y+z)^{2}$ , and noting the conclusion of problem 3: $a b+b c+c a \leqslant 3$ , we get
$\begin{array}{l} a b c(a+b+c)=a b \cdot b c+b c \cdot c a+c a \cdot a b \\ \leqslant \frac{1}{3}(a b+b c+c a)^{2} \\ \leqslant a b+b c+c a \end{array}$

Thus, $a b c(a+b+c) \leqslant a b+b c+c a$ ,
Therefore, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant a+b+c$ .

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