Triangle inequality square #2

Question

Let $a,b \in [0,1], a+b=1$ and let $\{u,v\}\subset\mathbb{R}$. How can we prove that $$(au+bv)^2\leq au^2+bv^2?$$

Opening the brackets give me in the left side $(au+v-av)^2$, but I'm not think that's the right way to do that.

I don't find the triangle inequality very helpfull, too.

Thanks.

Answer 1

Let $f(x)=x^2$.

Thus, $f$ is a convex function and your inequality it's just the Jensen inequality for $f$.

Also, C-S helps: $$au^2+bv^2=(a+b)(au^2+bv^2)\geq(au+bv)^2$$

Answer 2

$0\leq (u-v)^2\Rightarrow $

$2uv\leq u^2+v^2\Rightarrow$

$2abuv\leq abu^2+abv^2\Rightarrow $

$a^2u^2+2abuv+b^2v^2\leq a^2 u^2+abu^2+abv^2+b^2v^2\Rightarrow$

$(au+bv)^2\leq(au)^2+ab(u^2+v^2)+(bv)^2=(a+b)(au^2+bv^2)=au^2+bv^2$