Prove $m\cos^2{\theta} + n\sin^2{\theta} < l \implies \sqrt{m}\cos^2{\theta} + \sqrt{n}\sin^2{\theta} < \sqrt{l} $ Prove that $m\cos^2{\theta} + n\sin^2{\theta} < l \implies \sqrt{m}\cos^2{\theta} + \sqrt{n}\sin^2{\theta} < \sqrt{l} $ for every $m, n, l >0$.

This problem can be found in 

LARSON, LOREN C., 1983: Problem-Solving Through Problems.
  Springer-Verlag, p. 255, 267.

The author proposes two solutions: the first one is based on the Cauchy-Schwarz inequality; the second one on the arithmetic-mean−geometric-mean inequality. A third solution, which uses the concavity of $ f(x) = \sqrt{x} $, is left as an exercise. 

I will write here the solution based on the second method: 
\begin{align*}  
(\sqrt{m}\cos^2{\theta} + \sqrt{n}\sin^2{\theta})^2 
&= m \cos^4{\theta} + 2 \sqrt{mn}\cos^2{\theta}\sin^2{\theta}+ n\sin^4{\theta} \\ 
& \leq m \cos^4{\theta} + (m + n)\cos^2{\theta}\sin^2{\theta}+ n\sin^4{\theta} \\ 
&= (m\cos^2{\theta} + n\sin^2{\theta})(\cos^2{\theta} + \sin^2{\theta}) \\ 
&= (m\cos^2{\theta} + n\sin^2{\theta})\\ 
&<l.
\end{align*}
