Subfield: Algebra


Question: Let $a, b$, and $c$ be positive real numbers satisfying $\min (a+b, b+c, c+a)>\sqrt{2}$ and $a^{2}+b^{2}+c^{2}=3$. Prove that

$$
\frac{a}{(b+c-a)^{2}}+\frac{b}{(c+a-b)^{2}}+\frac{c}{(a+b-c)^{2}} \geq \frac{3}{(a b c)^{2}}
\tag{1}
$$

Throughout both solutions, we denote the sums of the form $f(a, b, c)+f(b, c, a)+f(c, a, b)$ by $\sum f(a, b, c)$.

Solution:
The condition $b+c>\sqrt{2}$ implies $b^{2}+c^{2}>1$, so $a^{2}=3-\left(b^{2}+c^{2}\right)<2$, i.e. $a<\sqrt{2}<b+c$. Hence we have $b+c-a>0$, and also $c+a-b>0$ and $a+b-c>0$ for similar reasons.

We will use the variant of HÖLDER's inequality

$$
\frac{x_{1}^{p+1}}{y_{1}^{p}}+\frac{x_{1}^{p+1}}{y_{1}^{p}}+\ldots+\frac{x_{n}^{p+1}}{y_{n}^{p}} \geq \frac{\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{p+1}}{\left(y_{1}+y_{2}+\ldots+y_{n}\right)^{p}}
$$

which holds for all positive real numbers $p, x_{1}, x_{2}, \ldots, x_{n}, y_{1}, y_{2}, \ldots, y_{n}$. Applying it to the left-hand side of (1) with $p=2$ and $n=3$, we get

$$
\sum \frac{a}{(b+c-a)^{2}}=\sum \frac{\left(a^{2}\right)^{3}}{a^{5}(b+c-a)^{2}} \geq \frac{\left(a^{2}+b^{2}+c^{2}\right)^{3}}{\left(\sum a^{5 / 2}(b+c-a)\right)^{2}}=\frac{27}{\left(\sum a^{5 / 2}(b+c-a)\right)^{2}}
\tag{2}
$$

To estimate the denominator of the right-hand part, we use an instance of ScHUR's inequality, namely

$$
\sum a^{3 / 2}(a-b)(a-c) \geq 0
$$

which can be rewritten as

$$
\sum a^{5 / 2}(b+c-a) \leq a b c(\sqrt{a}+\sqrt{b}+\sqrt{c})
$$

Moreover, by the inequality between the arithmetic mean and the fourth power mean we also have

$$
\left(\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{3}\right)^{4} \leq \frac{a^{2}+b^{2}+c^{2}}{3}=1
$$

i.e., $\sqrt{a}+\sqrt{b}+\sqrt{c} \leq 3$. Hence, (2) yields

$$
\sum \frac{a}{(b+c-a)^{2}} \geq \frac{27}{(a b c(\sqrt{a}+\sqrt{b}+\sqrt{c}))^{2}} \geq \frac{3}{a^{2} b^{2} c^{2}}
$$

thus solving the problem.
we mention that all the numbers $b+c-a, a+c-b, a+b-c$ are positive. We will use only this restriction and the condition

$$
a^{5}+b^{5}+c^{5} \geq 3
\tag{3}
$$

which is weaker than the given one. Due to the symmetry we may assume that $a \geq b \geq c$.

In view of (3), it suffices to prove the inequality

$$
\sum \frac{a^{3} b^{2} c^{2}}{(b+c-a)^{2}} \geq \sum a^{5}
$$

or, moving all the terms into the left-hand part,

$$
\sum \frac{a^{3}}{(b+c-a)^{2}}\left((b c)^{2}-(a(b+c-a))^{2}\right) \geq 0
\tag{4}
$$

Note that the signs of the expressions $(y z)^{2}-(x(y+z-x))^{2}$ and $y z-x(y+z-x)=(x-y)(x-z)$ are the same for every positive $x, y, z$ satisfying the triangle inequality. So the terms in (4) corresponding to $a$ and $c$ are nonnegative, and hence it is sufficient to prove that the sum of the terms corresponding to $a$ and $b$ is nonnegative. Equivalently, we need the relation

$$
\frac{a^{3}}{(b+c-a)^{2}}(a-b)(a-c)(b c+a(b+c-a)) \geq \frac{b^{3}}{(a+c-b)^{2}}(a-b)(b-c)(a c+b(a+c-b)) .
$$

Obviously, we have

$$
a^{3} \geq b^{3} \geq 0, \quad 0<b+c-a \leq a+c-b, \quad \text { and } \quad a-c \geq b-c \geq 0
$$

hence it suffices to prove that

$$
\frac{a b+a c+b c-a^{2}}{b+c-a} \geq \frac{a b+a c+b c-b^{2}}{c+a-b}
$$



Since all the denominators are positive, it is equivalent to

$$
(c+a-b)\left(a b+a c+b c-a^{2}\right)-\left(a b+a c+b c-b^{2}\right)(b+c-a) \geq 0
$$

or

$$
(a-b)\left(2 a b-a^{2}-b^{2}+a c+b c\right) \geq 0
$$

Since $a \geq b$, the last inequality follows from

$$
c(a+b)>(a-b)^{2}
$$

which holds since $c>a-b \geq 0$ and $a+b>a-b \geq 0$.