1976_Question_2

This question is trivial by Nesbitt’s Inequality so I shall prove Nesbitt’s inequality.
Nesbitt’s Inequality:
For positive a,b,c,
\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}

Proof:
By Cauchy-Schwarz we get that
[(b+c)+(a+c)+(a+b)][(\frac{1}{b+c})+(\frac{1}{a+c})+(\frac{1}{a+b})] \geq 9

This simplifies to 2(\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}) \geq 9.

Now we can take out the \frac{b+c}{b+c}, \frac{a+c}{a+c}, and \frac{a+b}{a+b} (cause they equal to 1).

Taking out the 3 and simplifying we get Nesbitt’s Inequality and equality holds when (b+c)^2=(a+c)^2=(a+b)^2 (from Cauchy-Schwarz).

This is just a=b=c and we can see that it does equal \frac{3}{2}.

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:star_struck: :star_struck: :star_struck: :star_struck: :star_struck: :star_struck: Nice!!!