## Saturday, June 11, 2016

### For positive reals $a,\,b,\,c$, prove that [MATH]\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\gt 2[/MATH].

For positive reals $a,\,b,\,c$, prove that [MATH]\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\gt 2[/MATH].

Hello all!

Today I'm going to post something that is going to be very different than my style in my previous blog posts, as today I wanted to train students to spot the factual mistake(s) that I might have or might not have made in the following solution (of mine) to today's delicious inequality problem.

Some of you might not aware that to be able to spot and tell and then rectify the mistakes of others or even to your own approach/strategy will further hence your performance, and to let people grow their confidence in you that you speak really well as a competitive students/worker. According to the billionaire Elon Musk, one of the most influential businessmen alive today, he quoted:

Being able to spot mistakes and communicate clearly on how to improve is a core managerial skill that will enhance employee performance efficiency.

Therefore, in light of this and in an effort to train students to spot the errors in the solutions of others, I have come to decide to do this blog post, and hopefully there are many of you who wanted to take a stab at it! :D

My solution:

First we multiply the top and bottom of each fraction by [MATH]\sqrt{a},\,\sqrt{b},\,\sqrt{c}[/MATH] respectively, and simplify along the way to get:

[MATH]\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}[/MATH]

[MATH]=\frac{\sqrt{a}}{\sqrt{b+c}}+\frac{\sqrt{b}}{\sqrt{c+a}}+\frac{\sqrt{c}}{\sqrt{a+b}}[/MATH]

[MATH]=\frac{\sqrt{a}\sqrt{a}}{\sqrt{a}\sqrt{b+c}}+\frac{\sqrt{b}\sqrt{b}}{\sqrt{b}\sqrt{c+a}}+\frac{\sqrt{c}\sqrt{c}}{\sqrt{c}\sqrt{a+b}}[/MATH]

[MATH]=\frac{a}{\sqrt{a}\sqrt{b+c}}+\frac{b}{\sqrt{b}\sqrt{c+a}}+\frac{c}{\sqrt{c}\sqrt{a+b}}[/MATH]

[MATH]\ge \frac{a}{\frac{a+b+c}{2}}+\frac{b}{\frac{a+b+c}{2}}+\frac{c}{\frac{a+b+c}{2}}[/MATH] (from the AM-GM inequality)

[MATH]\ge 2\left(\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}\right)[/MATH]

[MATH]= 2\left(\frac{a+b+c}{a+b+c}\right)[/MATH]

[MATH]= 2 [/MATH](Q.E.D.)