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

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.

• Does Cauchy-Schwarz produce any results? Apr 29, 2020 at 7:52
• Were I to use the Cauchy-Schwarz inequality, it would be $$\left(\frac{\sqrt m}{a} + \frac{\sqrt n}{b} + \frac{\sqrt p}{c}\right)^2 \le (m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$, if that was what you had meant. Apr 29, 2020 at 10:04
• @Lê Thành Đạt I proved your inequality for any reals $a$, $b$ and $c$ such that $abc\neq0$. Which says that it's proved for $a=mid\{a,b,c\}$. Jun 11, 2020 at 19:38
• Mr. @MichaelRozenberg , I'm doing a research on it.. I think you'll have such a beautiful solution for my favor !
– user552223
Jun 15, 2020 at 10:48

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.

• @Lê Thành Đạt I added something. See now. Apr 30, 2020 at 4:44
• @MichaelRozenberg (+1) It is a nice solution. Apr 30, 2020 at 5:51
• @MichaelRozenberg A minor question: "which happens if", should it be "which happens only if", or "which implies"? Apr 30, 2020 at 14:33
• Which gives, is it OK? Apr 30, 2020 at 14:51
• @MichaelRozenberg "which gives" is fine. Apr 30, 2020 at 15:33

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.

• (+1) Nice SOS (Sum of Squares). May 2, 2020 at 10:10
• @River Li Thanks. May 2, 2020 at 11:31
• Mr. @knvy144444 , will you have a fresh solution for $c:={\rm mid}\{a, b, c\}$ for my own problem math.stackexchange.com/q/3138409/552223 ?
– user552223
Jun 16, 2020 at 3:45

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.

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 $$[ab(a+b)+bc(c+a)+ca(c+a)]^2-4(ab+bc+ca)(a^2b^2+b^2c^2+c^2a^2)$$ $$=\sum a^2b^2(a-b)^2 + \sum 2abc \sum a(a-b)(a-c).$$

• Mr. @Nguyenhuyen_AG , do you want to help me the favor and receive my bounty ?
– user552223
Jun 15, 2020 at 10:57
• (+1) It is a nice solution. Jun 15, 2020 at 11:47

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 $$9+2x+x^2=\frac{4abc}{(a+b+c)^3}x^3.$$