Assume$ \sum_{i=1}^{n} x_i^2=1 $ and $ \sum_{i=1}^{n} y_i^2=1 $ $ \left(x_i,y_i>0,\forall i=1,2,\dots,n\right) $, denote $ a=\sum_{i=1}^{n} x_iy_i $
Ask to prove $$ \sum_{i=1}^{n}x_i^2\ln x_i+\sum_{i=1}^{n}y_i^2\ln y_i\leq a^2\sum_{i=1}^{n}\frac{x_iy_i}{a}\ln \frac{x_iy_i}{a} $$
Disgusting. My first attempt is to make it $$ \sum_{i=1}^{n}x_i^2\ln x_i^2+\sum_{i=1}^{n}y_i^2\ln y_i^2\leq 2a^2\sum_{i=1}^{n}\frac{x_iy_i}{a}\ln \frac{x_iy_i}{a} $$
No further results.
Another idea I thought about is to apply Jensen's Inequality upon function$ f(x)=x\ln x $
No further results either. Utterly disgusted.
Any Advice or Idea Would Be Greatly Appreciated!
