Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 \ge a + b + c$.

Question

Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 \ge a + b + c$.

I'm supposed to prove this use AM-GM, but can't figure it out. Any hints?

Answer 1

since $$a^2+1\ge 2a$$ so $$a^2+b^2+c^2+3\ge a+b+c+(a+b+c)\ge a+b+c+3\sqrt[3]{abc}=a+b+c+3$$ so $$a^2+b^2+c^2\ge a+b+c$$

Answer 2

Solution from page 5 in Thomas Mildorf's notes: first we homogenize the inequality with multipliying $(abc)^\frac{1}{3}$ and we get the following inequality $$a^\frac{4}{3}b^\frac{1}{3}c^\frac{1}{3}+a^\frac{1}{3}b^\frac{4}{3}c^\frac{1}{3}+a^\frac{1}{3}b^\frac{1}{3}c^\frac{4}{3}\:\leqslant\: a^2+b^2+c^2\,,$$ then we can apply $GM\leqslant AM\,$ to obtain $$a^\frac43b^\frac13 c^\frac13 \:\leqslant\: \frac{a^2+a^2+a^2+a^2+b^2+c^2}6\,.$$ And we can get in the same manner the other terms, and after adding them we are done.

Answer 3

By QM-AM for the first inequality and AM-GM for the second,

$$\frac{a^2+b^2+c^2}{3}\geq\left( \frac{a+b+c}{3} \right)^2\geq\frac{a+b+c}{3}\left( \sqrt[3]{abc} \right)$$

so since $abc=1, a^2+b^2+c^2\geq a+b+c$

Answer 4

Using C-S and then AM-GM gives something shorter: $$3(a^2+b^2+c^2) \ge (a+b+c)^2 \Rightarrow a^2+b^2+c^2 \ge (a+b+c)\frac{(a+b+c)}{3} $$ $$\ge (a+b+c)\sqrt[3]{abc}=a+b+c$$ And we're done.