Subfield: Algebra


Question: Let $a, b, c$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that

$$
\frac{1}{(2 a+b+c)^{2}}+\frac{1}{(2 b+c+a)^{2}}+\frac{1}{(2 c+a+b)^{2}} \leq \frac{3}{16} .
$$

Solution:
For positive real numbers $x, y, z$, from the arithmetic-geometric-mean inequality,

$$
2 x+y+z=(x+y)+(x+z) \geq 2 \sqrt{(x+y)(x+z)}
$$

we obtain

$$
\frac{1}{(2 x+y+z)^{2}} \leq \frac{1}{4(x+y)(x+z)}
$$

Applying this to the left-hand side terms of the inequality to prove, we get

$$
\begin{aligned}
\frac{1}{(2 a+b+c)^{2}} & +\frac{1}{(2 b+c+a)^{2}}+\frac{1}{(2 c+a+b)^{2}} \\
& \leq \frac{1}{4(a+b)(a+c)}+\frac{1}{4(b+c)(b+a)}+\frac{1}{4(c+a)(c+b)} \\
& =\frac{(b+c)+(c+a)+(a+b)}{4(a+b)(b+c)(c+a)}=\frac{a+b+c}{2(a+b)(b+c)(c+a)} .
\end{aligned}
\tag{1}
$$

A second application of the inequality of the arithmetic-geometric mean yields

$$
a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b \geq 6 a b c
$$

or, equivalently,

$$
9(a+b)(b+c)(c+a) \geq 8(a+b+c)(a b+b c+c a)
\tag{2}
$$

The supposition $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$ can be written as

$$
a b+b c+c a=a b c(a+b+c) .
\tag{3}
$$

Applying the arithmetic-geometric-mean inequality $x^{2} y^{2}+x^{2} z^{2} \geq 2 x^{2} y z$ thrice, we get

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

which is equivalent to

$$
(a b+b c+c a)^{2} \geq 3 a b c(a+b+c) .
\tag{4}
$$



Combining (1), (2), (3), and (4), we will finish the proof:

$$
\begin{aligned}
\frac{a+b+c}{2(a+b)(b+c)(c+a)} & =\frac{(a+b+c)(a b+b c+c a)}{2(a+b)(b+c)(c+a)} \cdot \frac{a b+b c+c a}{a b c(a+b+c)} \cdot \frac{a b c(a+b+c)}{(a b+b c+c a)^{2}} \\
& \leq \frac{9}{2 \cdot 8} \cdot 1 \cdot \frac{1}{3}=\frac{3}{16}
\end{aligned}
$$
Equivalently, we prove the homogenized inequality

$$
\frac{(a+b+c)^{2}}{(2 a+b+c)^{2}}+\frac{(a+b+c)^{2}}{(a+2 b+c)^{2}}+\frac{(a+b+c)^{2}}{(a+b+2 c)^{2}} \leq \frac{3}{16}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$$

for all positive real numbers $a, b, c$. Without loss of generality we choose $a+b+c=1$. Thus, the problem is equivalent to prove for all $a, b, c>0$, fulfilling this condition, the inequality

$$
\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}} \leq \frac{3}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
\tag{5}
$$

Applying JENSEN's inequality to the function $f(x)=\frac{x}{(1+x)^{2}}$, which is concave for $0 \leq x \leq 2$ and increasing for $0 \leq x \leq 1$, we obtain

$$
\alpha \frac{a}{(1+a)^{2}}+\beta \frac{b}{(1+b)^{2}}+\gamma \frac{c}{(1+c)^{2}} \leq(\alpha+\beta+\gamma) \frac{A}{(1+A)^{2}}, \quad \text { where } \quad A=\frac{\alpha a+\beta b+\gamma c}{\alpha+\beta+\gamma}
$$

Choosing $\alpha=\frac{1}{a}, \beta=\frac{1}{b}$, and $\gamma=\frac{1}{c}$, we can apply the harmonic-arithmetic-mean inequality

$$
A=\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \leq \frac{a+b+c}{3}=\frac{1}{3}<1
$$

Finally we prove (5):

$$
\begin{aligned}
\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}} & \leq\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \frac{A}{(1+A)^{2}} \\
& \leq\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \frac{\frac{1}{3}}{\left(1+\frac{1}{3}\right)^{2}}=\frac{3}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) .
\end{aligned}
$$