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

Solution 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 $$a^2+b^2+c^2\ge a+b+c$$

Solution 2:

Solution from Thomas Mildorf notes first we homogenize inequality with multipliying $(abc)^\frac{1}{3}$ and we get the following inequality $$a^\frac{4}{3}b^\frac{1}{3}c^\frac{1}{3}+a^\frac{1}{3}b^\frac{4}{3}c^\frac{1}{3}+a^\frac{1}{3}b^\frac{1}{3}c^\frac{4}{3}\leq a^2+b^2+c^2$$ then we can apply $A.O\geq H.O$ $$\frac{a^2+a^2+a^2+a^2+b^2+c^2}{6}\geq a^\frac{4}{3}b^\frac{1}{3}c^\frac{1}{3}$$ and we can get in the same manner other term and after adding them we are done

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

Solution 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.