If $a,b,c > 0$ prove $\frac{(b+c-a)^2}{a}+\frac{(a+c-b)^2}{b}+\frac{(a+b-c)^2}{c}\ge\frac{b^2+c^2}{b+c}+\frac{a^2+c^2}{a+c}+\frac{a^2+b^2}{a+b}$

Question

$a,b,c > 0$ $$\frac{(b+c-a)^2}{a}+\frac{(a+c-b)^2}{b}+\frac{(a+b-c)^2}{c}\ge\frac{b^2+c^2}{b+c}+\frac{a^2+c^2}{a+c}+\frac{a^2+b^2}{a+b}$$

I have used Tittu here with the second and the third fraction and came with this result :

$$\frac{(a+c-b)^2}{b}+\frac{(a+b-c)^2}{c}\ge\frac{(a+c-b+a+b-c)^2}{b+c}=\frac{4a^2}{b+c}$$

If you do this (taking fractions two by two) then you have to divide them by 2 in order to get the LHS and thus : $$LHS \ge \frac{2a^2}{b+c}+\frac{2b^2}{a+c}+\frac{2c^2}{a+b}$$

I believe I am quite near solving this but I can't come up with something. If you can, please help and thank you.

Answer 1

Let's try to complete from OP's approach. Notice that in order to apply Titu's lemma, we must check that the terms are positive, but they need not be in this setup. We'd use a variant of Titu's lemma, which is an immediate corollary of Titu's lemma.

For real (not necessarily positive) numbers $a_i$ and positive reals $b_i$, we have
$$\sum \frac{ a_i^2 } { b_i } \geq \frac { (\sum a_i)^2 } { b_i }. $$
Proof: Titu's Lemma gives us $\sum \frac{ a_i^2 } { b_i } \geq \frac { (\sum |a_i|)^2 } { \sum b_i } $, and it's obvious that $\frac { (\sum |a_i|)^2 } { \sum b_i } \geq \frac{ (\sum a_i)^2 } { \sum b_i}$.

Applying this, we may correctly conclude as OP did that

$$LHS \geq \sum \frac{ 2a^2 } { b+c}.$$

Then, $ \{a^2, b^2, c^2 \}$ is similarly ordered with $ \{a, b, c \}$, oppositely ordered with $ \{ (a+b+c) -a, (a+b+c) - b, (a+b+c) - c \} $, and hence similarly ordered with $ \{ \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \}$. Hence by the rearrangement inequality, the product is maximized when we take $(a^2, b^2, c^2 ) \cdot ( \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b})$ vs any other arrangement, thus we get that $ \sum \frac{a^2}{b+c} \geq \sum \frac{b^2}{b+c}$ and also $ \sum \frac{a^2}{b+c} \geq \sum \frac{b^2}{b+c}$.

Hence $ LHS \geq \sum \frac{ 2a^2}{b+c} \geq \sum \frac{ b^2+c^2}{b+c} $.

Equality holds iff $a=b=c$.

Answer 2

We need to prove that: $$\sum_{cyc}\left(\frac{(b+c-a)^2}{a}-a\right)\geq\sum_{cyc}\left(\frac{a^2+b^2}{a+b}-\frac{a+b}{2}\right)$$ or $$\sum_{cyc}\frac{(b+c-2a)(b+c)}{a}\geq\sum_{cyc}\frac{(a-b)^2}{2(a+b)}$$ or $$\sum_{cyc}\frac{(c-a-(a-b))(b+c)}{a}\geq\sum_{cyc}\frac{(a-b)^2}{2(a+b)}$$ or $$\sum_{cyc}(a-b)\left(\frac{a+c}{b}-\frac{b+c}{a}\right)\geq\sum_{cyc}\frac{(a-b)^2}{2(a+b)}$$ or $$\sum_{cyc}\frac{(a-b)^2(a+b+c)}{ab}\geq\sum_{cyc}\frac{(a-b)^2}{2(a+b)}$$ or $$\sum_{cyc}(a-b)^2\left(\frac{a+b+c}{ab}-\frac{1}{2(a+b)}\right)\geq0,$$ which is obvious.