Prove $\sqrt{a} \cos^2(\theta) + \sqrt{b} \sin^2(\theta) < \sqrt{c}$ if $a\cos^2(\theta) + b \sin^2(\theta) < c$

Question

Use the concavity of $f(x) = \sqrt{x}$ to prove that if $a, b,c$ are positive, then $a\cos^2(\theta) + b \sin^2(\theta) < c$ implies $\sqrt{a} \cos^2(\theta) + \sqrt{b} \sin^2(\theta) < \sqrt{c}$

I can prove this using Cauchy-Schwarz, but I am struggling to prove it using concavity of $\sqrt{x}$. Could I get a hint? Please $\textbf{only post hints}$.

Concave means $f''(x)<0$

Answer 1

Hint

Apply concave inequality $$f(tx+(1-t)y)\ge tf(x)+(1-t)f(y)$$ with $t=\cos^2 \theta$ and$x,y$ well (and easily) chosen.

Answer 2

Hint (for another way, not concavity involved)

Define $f(\theta)= a\cos^2(\theta) + b \sin^2(\theta) $ and $g(\theta)= \sqrt{a}\cos^2(\theta) + \sqrt{b}\sin^2(\theta) $.

So taking derivative for the first function and applying the $0$-derivative theorem you obtain that:

$$\max{(f(\theta))}= \max{(a,b)}$$$$\min{(f(\theta))}= \min{(a,b)} $$

But then you can get easily max and min of $g(\theta)$ and you know also $ \max{(f(\theta))} < c $ , so...