proof of inequality-using a convex function

Question

Let $a_{1},a_{2},\ldots,a_{n},b_{1},b_{2},\ldots,b_{n}$ be positive numbers. We need to prove that: $$(a_{1}+b_{1})^{\frac{1}{n}}(a_{2}+b_{2})^{\frac{1}{n}}\cdots(a_{n}+b_{n})^{\frac{1}{n}}\geq a_{1}^{\frac{1}{n}}a_{2}^{\frac{1}{n}} a_{n}^{\frac{1}{n}}+b_{1}^{\frac{1}{n}}b_{2}^{\frac{1}{n}}\cdots b_{n}^{\frac{1}{n}}$$

I came up with a proof for this problem by simply using Arithmetic-Geometric Mean Inequality as follows:

$$ \frac{a_{1}^{\frac{1}{n}}a_{2}^{\frac{2}{n}}\cdots a_{n}^{\frac{1}{n}}+b_{1}^{\frac{1}{n}}b_{2}^{\frac{1}{n}}\cdots b_{n}^{\frac{1}{n}} }{(a_{1}+b_{1})^{\frac{1}{n}}(a_{2}+b_{2})^{\frac{1}{n}}\cdots (a_{n}+b_{n})^{\frac{1}{n}}}=(\frac{a_{1}}{a_{1}+b_{1}})^{\frac{1}{n}}(\frac{a_{2}}{a_{2}+b_{2}})^{\frac{1}{n}}\cdots(\frac{a_{n}}{a_{n}+b_{n}})^{\frac{1}{n}}+(\frac{b_{1}}{a_{1}+b_{1}})^{\frac{1}{n}}(\frac{b_{2}}{a_{2}+b_{2}})^{\frac{1}{n}}\cdots(\frac{b_{n}}{a_{n}+b_{n}})^{\frac{1}{n}}=\sqrt[n]{(\frac{a_{1}}{a_{1}+b_{1}})(\frac{a_{2}}{a_{2}+b_{2}})\cdots(\frac{a_{n}}{a_{n}+b_{n}})}+\sqrt[n]{(\frac{b_{1}}{a_{1}+b_{1}})(\frac{b_{2}}{a_{2}+b_{2}})\cdots(\frac{b_{n}}{a_{n}+b_{n}})}\leq \frac{1}{n}\left [ (\frac{a_{1}}{a_{1}+b_{1}})+(\frac{a_{2}}{a_{2}+b_{2}})+\cdots+(\frac{a_{n}}{a_{n}+b_{n}}) \right ]+\frac{1}{n}\left [(\frac{b_{1}}{a_{1}+b_{1}})+ (\frac{b_{2}}{a_{2}+b_{2}})+\cdots+(\frac{b_{n}}{a_{n}+b_{n}}) \right ]=1$$

where last inequality follows from applying the Arithmetic-Geometric Mean inequality.

The professor said there is a short way to solve the problem by proving that the function (I don't know which function he was talking about) is convex, and then the inequality follows immediately. I am really interested to know this method, so I appreciate if someone shares it with me.

Answer 1

The proof by AM-GM inequality is indeed elegant, you can see it again here.

In Exercise 6.5 of J. Michael Steele's The Cauchy-Schwarz Master Class, you are asked to find an alternative proof using Jensen's inequality. Here is Steele's solution:

To build a proof with Jensen's inequality, we first divide by $(a_1a_2\cdots a_n)^{1/n}$ and write $c_k$ for $b_k/a_k$, so the target inequality takes the form $$1+(c_1c_2\cdots c_n)^{1/n}\leq \left\{(1+c_1)(1+c_2)\cdots(1+c_n)\right\}^{1/n}. $$ Now if we take logs and write $c_j$ as $\exp(d_j)$, we find it takes the form $$\log(1+\exp(\bar d))\leq {1\over n}\sum_{j=1}^n \log(1+\exp(d_j)), $$ where $\bar d=(d_1+d_2+\cdots +d_n)/n$. Finally, the last inequality is simply Jensen's inequality for the convex function $x\mapsto \log(1+e^x)$, so the solution is complete.