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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 Cauchy-Schwarz inequality.

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.