Prove that for all positive real numbers $a,b,c$ the following inequality holds: \[(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\ge\frac{2(a^2+b^2+c^2)}{ab+bc+ca}+7\]and determine all cases of equality. Lucian Petrescu
Problem
Source: Romanian NMO 2021 grade 8 P2
Tags: inequalities
arqady
15.04.2023 16:58
Miquel-point wrote: Prove that for all positive real numbers $a,b,c$ the following inequality holds: \[(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\ge\frac{2(a^2+b^2+c^2)}{ab+bc+ca}+7\]and determine all cases of equality. Lucian Petrescu By uvw it's enough to check one case only: $b=c=1$.
sqing
15.04.2023 17:05
Prove that for all positive real numbers $a,b,c$ the following inequality holds: \[(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\ge\frac{4(a^2+b^2+c^2)}{ab+bc+ca}+5\]
S.Lenny
15.04.2023 17:17
Subtracting 9 from both sides the original inequality is equivalent to $\frac{(a-b)^2}{ab}+\frac{(b-c)^2}{bc}+\frac{(c-a)^2}{ca} \ge \frac{(a-b)^2+(b-c)^2+(c-a)^2}{ab+bc+ca}$ Which is obvious.
DensSv
12.01.2025 16:47
After some trivial calculations, the inequality is equivalent to
$$\sum \frac{a+b}{c}\geq \frac{2(a+b+c)^{2}}{ab+bc+ca}$$However by the Mircea Lascu inequality, which states that
$$\sum \frac{a+b}{c}\geq \sum\frac{4a}{b+c}$$and it's demonstration relies on the fact that $\frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}$ and multiply the inequality with $c$ and sum up their reciprocals.
$$\sum \frac{4a}{b+c}=\sum \frac{4a^{2}}{ab+ac}\geq \frac{4(a+b+c)^{2}}{2(ab+bc+ca)}=\frac{2(a+b+c)^{2}}{ab+bc+ca}$$Which is what we wanted to prove.