Proving inequality $x\ln(x)+y\ln(y)\geq(x+y)\ln(\frac{x+y}{2})$

Question

How can we prove that for $x,y>0$ we have $$ x\ln(x)+y\ln(y)\geq(x+y)\ln\left(\frac{x+y}{2}\right) $$ using the derivatives of functions of one variable?

Answer 1

Hint: This is same as: $$\frac{\ln{x^x}+\ln{y^y}}{2}\geq \ln{(\frac{x+y}{2})^{(x+y)/2}}$$

Consider the function $f(k)=\ln{k^k}$, $k>0$. The above inequality means the midpoint of the line joining the points $f(x)$ and $f(y)$ is above $f(\frac{x+y}{2})$. Which is true for any convex function. So you just need to prove that the function is convex. Can you continue now?

Answer 2

WLOG $y\ge x$ and by setting $y=xt$, the relation simplifies to $$ t\ln t\geq(1+t)\ln\frac{1+t}2. $$

Notice that the two members coincide at $t=1$, so that we may take the derivative of the inequality,

$$\ln t+1\ge \ln\frac{1+t}2+1$$

which is just $$t\ge\frac{1+t}2.$$

Answer 3

If $x=y$ we are done. WOLG we can assume that $x \ge y$. For $t \ge y$ define

$f(t)=t \ln t+y \ln y-(t+y) \ln (\frac{t+y}{2}).$ Then $f(y)=0$. Now let $x>y$. By the mean value theorem there is $s \in (y,x)$ such that

$$f(x)=f(x)-f(y)= f'(s)(x-y).$$

It is easy to see that $f'(s) \ge 0$, hence, since $x-y >0$, we derive $f(x) \ge 0.$

Answer 4

Let $f(t)=(t+1)\ln(t+1)-t\ln t-(t+1)\ln 2$ for $t>0$. Then

$$f'(t)=1+\ln(t+1)-1-\ln t-\ln 2\begin{cases}>0\quad\textrm{if }t\in(0,1)\\=0\quad\textrm{if }t=1\\<0\quad\textrm{if }t\in(1,\infty)\end{cases}$$

$f$ attains its maximum at $t=1$.

$$(t+1)\ln(t+1)-t\ln t-(t+1)\ln 2\le2\ln2-\ln1-2\ln 2=0$$

Let $\displaystyle t=\frac{x}{y}$. We have

$$\left(\frac{x}{y}+1\right)\ln\left(\frac{x}{y}+1\right)\le\frac{x}{y}\ln \left(\frac{x}{y}\right)+\left(\frac{x}{y}+1\right)\ln 2$$

$$(x+y)\ln\left(\frac{x+y}{y}\right)\le x\ln\left(\frac{x}{y}\right)+(x+y)\ln2$$

$$(x+y)\ln\left(\frac{x+y}{2}\right)-(x+y)\ln y\le x\ln(x)-x\ln(y)$$

$$(x+y)\ln\left(\frac{x+y}{2}\right)\le x\ln(x)+y\ln(y)$$