Let $a,\,b$ and $c$ be positive real that is greater than $1$ such that [MATH]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2[/MATH].

Prove that $\sqrt{a+b+c}\ge \sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}$.

My solution:

From the given equality, we can algebraically manipulate it so that it turns to become an useful "weapon" to help us to prove the intended inequality...

But, there are plenty of algebraic tricks we could employ to turn the equality to take different multiple forms...take for examples, we could get some of the following:

i. [MATH]\frac{bc}{abc}+\frac{ac}{abc}+\frac{ab}{abc}=2\implies ab+bc+ca=2abc[/MATH]

ii. [MATH]\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}=1[/MATH]

The list could go on...we therefore should be super clear of our objective, otherwise, we could end up running in circle and going nowhere nearer to the goal! Our aim is, we need to prove the LHS of the inequality greater than or equal to the sum of the three radical terms where their arguments are $a-1,\,b-1,\,c-1$.

That gives us pretty much hint that we know we should get an equality with those $a-1,\,b-1,\,c-1$ in it...

How? We could, however, minus 3 from both sides of the equality and get:

[MATH]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-3=2-3[/MATH]

[MATH]\frac{1}{a}-1+\frac{1}{b}-1+\frac{1}{c}-1=-1[/MATH]

[MATH]\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}-1=-1[/MATH]

[MATH]\color{yellow}\bbox[5px,purple]{\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-1}{c}-1=1}[/MATH]

Now, with the help of the Cauchy-Schwarz inequality, we can start prove the intended inequality:

$\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}$

$=\frac{\sqrt{a}}{\sqrt{a}}\cdot \sqrt{a-1}+\frac{\sqrt{b}}{\sqrt{b}}\cdot \sqrt{b-1}+\frac{\sqrt{c}}{\sqrt{c}}\cdot \sqrt{c-1}$

$=\sqrt{a}\cdot \frac{\sqrt{a-1}}{\sqrt{a}}+\sqrt{b}\cdot \frac{\sqrt{b-1}}{\sqrt{b}}+\sqrt{c}\cdot \frac{\sqrt{c-1}}{\sqrt{c}}$

$\le \sqrt{a+b+c}\sqrt{\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-1}{c}}$ (By Cauchy-Schwarz Inequality)

$\le \sqrt{a+b+c}\sqrt{1}$ (from [MATH]\color{yellow}\bbox[5px,purple]{\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-1}{c}-1=1}[/MATH])

$\le \sqrt{a+b+c}$ (Q.E.D.)

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