Prove that for all positives $a, b$ and $c$, $(\sum_{cyc}\frac{c + a}{b})^2 \ge 4(\sum_{cyc}ca)(\sum_{cyc}\frac{1}{b^2})$.

Question

Prove that for all positives $a, b$ and $c$, $$\left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)^2 \ge 4(bc + ca + ab) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$

Let $ca + ab = m$, $ab + bc = n$ and $bc + ca = p$, we have that $$\left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2}\right)^2 \ge 2(m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$

$$\iff \left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2} - 1\right)^2 \ge 2 \cdot \sum_{cyc((m, n, p), (a, b, c))}\left[n \cdot \left(\frac{1}{c^2} + \frac{1}{a^2}\right)\right] + 1$$

Expanding $\displaystyle \sum_{cyc((m, n, p), (a, b, c))}\left[n \cdot \left(\frac{1}{c^2} + \frac{1}{a^2}\right)\right]$ gives $$2 \cdot \sum_{cyc}\frac{ca}{b^2} + \left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)$$

Let $\dfrac{b + c}{a} = m'$, $\dfrac{c + a}{b} = n'$ and $\dfrac{a + b}{c} = p'$, we have that $$(m' + n' + p' - 1)^2 \ge 2 \cdot \left[2 \cdot \sum_{cyc}\frac{ca}{b^2} + (m' + n' + p')\right] + 1$$

Moreover, $$(m')^2 + (n')^2 + (p')^2 = \sum_{cyc}\left[\left(\frac{c + a}{b}\right)^2\right] \ge 2 \cdot \sum_{cyc}\frac{ca}{b^2}$$

$$\implies (m' + n' + p' - 1)^2 \ge 2 \cdot \left[(m')^2 + (n')^2 + (p')^2 + m' + n' + p'\right] + 1$$

$$\iff -[(m')^2 + (n')^2 + (p')^2] + 2(m'n' + n'p' + p'm') - 4(m' + n' + p') \ge 0$$, which is definitely not correct.

Another attempt, let $(0 <) \ a \le b \le c \implies ab \le ca \le bc \iff ca + ab \le ab + bc \le bc + ca$

$\iff m \le n \le p$ and $a^2 \le b^2 \le c^2 \iff \dfrac{1}{a^2} \ge \dfrac{1}{b^2} \ge \dfrac{1}{c^2}$.

By the Chebyshev inequality, we have that $$3 \cdot \left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2}\right) \le (m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$

Any help would be appreciated.

Answer 1

Let $a+b+c=x(a^2+b^2+c^2).$

Thus, $$\sum_{cyc}(a-xa^2)^2\geq0=\left(\sum_{cyc}(a-xa^2)\right)^2$$ or $$\sum_{cyc}(a^2-2xa^3+x^2a^4)\geq\sum_{cyc}(a^2+2ab)-2x\sum_{cyc}(a^3+a^2b+a^2c)+x^2\sum_{cyc}(a^4+2a^2b^2)$$ or $$\sum_{cyc}ab-x\sum_{cyc}(a^2b+a^2c)+x^2\sum_{cyc}a^2b^2\leq0.$$ Now, let $a^2b^2+a^2c^2+b^2c^2=A$, $\sum\limits_{cyc}ab(a+b)=B$ and $ab+ac+bc=C$.

We proved that there is $x$, id est, $x=\frac{a+b+c}{a^2+b^2+c^2}$ for which $A>0$ and $$Ax^2-Bx+C\leq0,$$ which gives $$\Delta\geq0$$ or $$B^2-4AC\geq0$$ or $$\left(\sum_{cyc}(a^2b+a^2c)\right)^2-4\sum_{cyc}ab\sum_{cyc}a^2b^2\geq0,$$ which is your inequality exactly.

Answer 2

WLOG assume $c\neq \text{mid}\{a,b,c\}$. We have:

$$a^2 b^2 c^2 (\text{LHS-RHS}) =\left( a-b \right) ^{2} \left( ab+ca+bc-{c}^{2} \right) ^{2}+4\,ab{c} ^{2} \left( a-c \right) \left( b-c \right) \geqq 0$$ Done.

Answer 3

We use the standard pqr method.

Let $p = a + b + c, q = ab+bc+ca, r = abc$. The inequality becomes $$\left(\frac{pq}{r} - 3\right)^2 \ge 4q \left(\left(\frac{q}{r}\right)^2 - 2\frac{p}{r}\right)$$ or (after clearing the denominators) $$p^2q^2 + 2pqr - 4q^3 + 9r^2 \ge 0.$$

We split into two cases:

1) $p^2 \ge 4q$: Since $p^2q^2 \ge 4q^3$, the inequality is true.

2) $p^2 < 4q$: By Schur's inequality $a^2(a-b)(a-c) + b^2(b-c)(b-a) + c^2(c-a)(c-b) \ge 0$ which is written as $6pr - (4q-p^2)(p^2-q) \ge 0$, we have $r \ge \frac{(4q-p^2)(p^2-q)}{6p}$. As $p^2 \ge 3q$, we have $r \ge \frac{(4q-p^2)(p^2-q)}{6p}\ge 0$. Thus, we have \begin{align} &p^2q^2 + 2pqr - 4q^3 + 9r^2\\ \ge\ & p^2q^2 + 2pq \cdot \frac{(4q-p^2)(p^2-q)}{6p} - 4q^3 + 9\left(\frac{(4q-p^2)(p^2-q)}{6p}\right)^2\\ = \ & \frac{(p^2-3q)(3p^2-q)(p^2-4q)^2}{12p^2}\\ \ge \ & 0. \end{align} We are done.

Answer 4

The inequality follows from the famous inequality Ukraine MO 2001.

If $ a,\,b,\,c$ and $x,\,y,\,z$ non-negative real numbers, then $$ [x(b+c) + y(c+a) + z(a+b)]^2 \geqslant 4(ab+bc+ca)(xy+yz+zx). \quad (1)$$ Now, using $(1)$ with $x=a^2,\,y=b^2,\,z=c^2$ we get $$ [a^2(b+c) + b^2(c+a) + c^2(a+b)]^2 \geqslant 4(ab+bc+ca)(a^2b^2+b^2c^2+c^2a^2),$$ or $$\left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)^2 \geqslant 4(ab+bc+ca)\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right).$$ Note. The sum of squares by Darij Grinberg $$[ab(a+b)+bc(c+a)+ca(c+a)]^2-4(ab+bc+ca)(a^2b^2+b^2c^2+c^2a^2)$$ $$=\sum a^2b^2(a-b)^2 + \sum 2abc \sum a(a-b)(a-c).$$

Answer 5

A NOTE

The given inequality $$ \left(\sum_{cyc}\frac{a+b}{c}\right)^2\geq 4\left(\sum_{cyc}ab\right)\left(\sum_{cyc}\frac{1}{a^2}\right)\textrm{, }a,b,c>0\tag 1 $$ is written as $$ \left(\sum_{cyc}ab(a+b)\right)^2\geq4\left(\sum_{cyc}ab\right)\left(\sum_{cyc}a^2b^2\right).\tag 2 $$ Hence if $s=a+b+c$ and $p=abc$ and $s_h=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, then $$ \left(\sum_{cyc}ab(a+b)\right)^2=p^2(ss_h-3)^2 $$ and $$ 4\left(\sum_{cyc}ab\right)\left(\sum_{cyc}a^2b^2\right)=4p^3s_h(s_h^2-2s/p) $$ Hence (2) becomes $$ 9+2ss_h+s^2s_h^2-4ps_h^3\geq 0.\tag 3 $$ Inequality (3) is a equivalent to $$ 9\leq ss_h\leq \rho\left(\frac{4p}{s^3}\right), $$ where $\rho=\rho(t)$ is the real root of $$ t \rho^3=9+2 \rho+\rho^2. $$ Hence given positive numbers $a,b,c$ we have $$ 9\leq (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\leq x, $$ where $x$ is the real root of $$ 9+2x+x^2=\frac{4abc}{(a+b+c)^3}x^3. $$