**Problem:** If \(a,b,c\) are positive numbers, then show that

\(\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}\geq a+b+c\)

**Solution:** This problem can be solved using a direct application of the * Titu’s Lemma* but we will instead prove the lemma first using the

*inequality.*

**Cauchy-Schwarz**According to the * Cauchy-Schwarz* inequality we have,

\(\left(a_1^2+a_2^2+\cdots+a_n^2\right)\left(b_1^2+b_2^2+\cdots+b_n^2\right)\ge \left(a_1b_1+a_2b_2+\cdots+a_nb_n\right)^2 \)

Replacing \(a_i\to \dfrac{a_i}{\sqrt{b_i}}\) and \(b_i\to \sqrt{b_i}\) we get,

\(\left(\dfrac{a_1^2}{b_1}+\dfrac{a_2^2}{b_2}+\cdots +\dfrac{a_n^2}{b_n}\right)\left(b_1+b_2+\cdots+b_n\right)\ge \left(a_1+a_2+\cdots+a_n\right)^2,\)

which is equivalent to

\(\dfrac{a_1^2}{b_1}+\dfrac{a_2^2}{b_2}+\cdots+\dfrac{a_n^2}{b_n}\geq \dfrac{\left(a_1+a_2+\cdots+a_n\right)^2}{b_1+b_2+\cdots+b_n}\)

Now this inequality is referred to as the **Titu’s Lemma.**

This brings us to the problem which can be observed to be a simple application of the lemma. We just need to make the following substitutions.

\(a_1=b, a_2=c\) and \(b_1=b_2=b+c\)

Then we have,

\(\dfrac{b^2}{b+c}+\dfrac{c^2}{b+c}\geq \dfrac{(b+c)^2}{2(b+c)}\)

\(=>\dfrac{b^2+c^2}{b+c}\geq \dfrac{b+c}{2}\)

Thus similarly we have,

\(=>\dfrac{c^2+a^2}{c+a}\geq \dfrac{c+a}{2}\) and \(=>\dfrac{a^2+b^2}{a+b}\geq \dfrac{a+b}{2}\)

Adding the three inequalities we get,

\(=>\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b}\geq \dfrac{b+c}{2}+\dfrac{c+a}{2}+\dfrac{a+b}{2}\)

\(=>\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b}\geq \dfrac{2(a+b+c)}{2}\)

\(=>\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b}\geq {a+b+c}\)

Hence Proved.

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