Is it true that $\sqrt{(a_1+b_1+c_1)(a_2+b_2+c_2)}\geq \sqrt{a_1a_2}+\sqrt{b_1b_2}+\sqrt{c_1c_2}$?

Question

Let $a_1,b_1,c_1,a_2,b_2,c_2$ be positive real numbers. Is it true that

$\sqrt{(a_1+b_1+c_1)(a_2+b_2+c_2)}\geq \sqrt{a_1a_2}+\sqrt{b_1b_2}+\sqrt{c_1c_2}$

If this is true, then I will have proved an Olympiad problem. I couldn't find a counter-example.

Answer 1

Square both sides (they are non-negative) and rewrite the terms: $$\iff(\sqrt{a_1}\sqrt{a_1}+\sqrt{b_1}\sqrt{b_1}+\sqrt{c_1}\sqrt{c_1})(\sqrt{a_2}\sqrt{a_2}+\sqrt{b_2}\sqrt{b_2}+\sqrt{c_2}\sqrt{c_2})\ge(\sqrt{a_1}\sqrt{a_2}+\sqrt{b_1}\sqrt{b_2}+\sqrt{c_1}\sqrt{c_2})^2$$ This is precisely the Cauchy–Schwarz inequality applied on $(\sqrt{a_1},\sqrt{b_1},\sqrt{c_1})$ and $(\sqrt{a_2},\sqrt{b_2},\sqrt{c_2})$.

Answer 2

An alternative proof (to the "standard" Cauchy-Schwarz one) is to verify that the inequality holds (as an equality) when all the variables equal $0$; and that, when all variables are non-negative, the partial derivative of the left-hand side with respect to any variable equals or exceeds the partial derivative of the right-hand side (so, if you "grow" the variables from $0$ to whatever non-negative values you want, the "advantage" of the left-hand side can only grow).

In fact, by the same argument if all variables are strictly positive, the inequality becomes strict!

Note that this argument does require some care if you want to formalize it, because the derivative of $\sqrt{x}$ with respect to $x$ diverges to $+\infty$ as $x$ converges to $0$.