Prove that for all positive real numbers $a, b, c$ with $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} =1$ the following inequality holds: $$3(ab+bc+ca)+\frac{9}{a+b+c} \le \frac{9abc}{a+b+c} + 2(a^2+b^2+c^2)+1$$
Problem
Source: Latvian TST for Baltic Way 2019 Problem 1
Tags: inequalities
Kimchiks926
29.05.2020 10:04
We multiply both sides of inequality by $(a+b+c)$. Then we need to prove that:
$$3(a+b+c)(ab+bc+ca)+9 \le 9abc +2(a^2+b^2+c^2)(a+b+c) + (a+b+c)$$After expanding we are left to prove:
$$\Sigma(a^2b+ab^2) + 9 \le 2(a^3+b^3+c^3) + (a+b+c) $$
Claim: $a^3 + b^3 \ge a^2b+ab^2$
Proof: Note that:
$$a^3+b^3 - ab(a+b) \ = (a+b)(a-b)^2 \ge 0 $$
From claim follows that $2(a^3+b^3+c^3) \ge \Sigma (a^2b+ ab^2) $
From Cauchy we obtain that:
$$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 9$$Since $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} =1$, we get that $a+b+c \ge 9$. So we conclude that:
$$\Sigma(a^2b+ab^2) + 9 \le 2(a^3+b^3+c^3) + (a+b+c) $$as desired
tzy
29.05.2020 10:41
This is great!
sqing
30.05.2020 14:36
Prove that for all positive real numbers $a, b$ with $\frac{1}{a}+\frac{1}{b}=1$ the following inequality holds: $$3ab+\frac{4}{a+b} \le a^2+b^2+5$$
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HamstPan38825
16.03.2023 02:15
Sum the two inequalities \begin{align*} \frac 9{a+b+c} &\leq 1 = \frac 1a+\frac 1b +\frac 1c \\ \frac{9abc}{a+b+c} + 2(a^2+b^2+c^2) &\geq 3(ab+bc+ca). \end{align*}The second inequality follows by multiplying both sides by $a+b+c$ and expanding, from where it becomes $$2(a^3+b^3+c^3) \geq \sum_{\mathrm{sym}} a^2b.$$
sqing
14.05.2023 05:30
Let $a, b, c, d$ be non-negative real numbers such that $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=3.$ Prove that \[3(ab+bc+ca+ad+bd+cd)+\frac{4}{a+b+c+d}\leq 5 \]2022 BMO Shortlist https://artofproblemsolving.com/community/c6h3071448p27726701 Let $a, b, c>0$. Prove that \[ \frac{a+3b}{b+c} +\frac{b+3c}{c+a}+\frac{c+3a}{a+b} \geq 6\]Let $a, b, c, d$ be positive numbers. Prove that $$\frac{a+3b}{b+c}+\frac{b+3c}{c+d}+\frac{c+3d}{d+a}+\frac{d+3a}{a+b}\geq8.$$
Saucepan_man02
14.05.2023 05:55
Kimchiks926 wrote: Prove that for all positive real numbers $a, b, c$ with $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} =1$ the following inequality holds: $$3(ab+bc+ca)+\frac{9}{a+b+c} \le \frac{9abc}{a+b+c} + 2(a^2+b^2+c^2)+1$$
Multiply both sides by $a+b+c$: $$9abc+2(a+b+c)(a^2+b^2+c^2)+a+b+c \ge 3(a+b+c)(ab+bc+ca)+9$$Expanding and cancelling common terms, we are left with: $$2(a^3+b^3+c^3)+a+b+c \ge (a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)+9$$From Muirhead, we have: $$a^3+b^3 \ge a^2b+ab^2$$Hence $$2(a^3+b^3+c^3) \ge (a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)$$Notice that, by AM-HM Inequality, we have: $$\frac{a+b+c}{3} \ge \frac{3}{ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}} = 3$$$$\implies a+b+c \ge 9$$Hence the above inequality holds. Equality holds when $a=b=c=3$.
dudade
22.06.2024 09:09
Note that by AM-HM, then \begin{align*} \dfrac{a + b + c}{3} \geq \dfrac{3}{\tfrac{1}{a} + \tfrac{1}{b} + \tfrac{1}{c}} = 3 \quad \longrightarrow \quad \dfrac{9}{a + b + c} \leq 1. \end{align*}Combining with $ab + bc + ca \leq a^2 + b^2 + c^2,$ we want to show \begin{align*} 3(ab+bc+ca) \le \frac{9abc}{a+b+c} + 2(a^2+b^2+c^2) \quad \longrightarrow \quad ab+bc+ca \leq \frac{9abc}{a+b+c}. \end{align*}The given condition implies $ab + bc + ca = abc$, but we know $a + b + c \leq 9$. Therefore, we are finished.
blueprimes
11.08.2024 04:08
who let them cook Let $a + b + c = 3u$, $ab + bc + ca = 3v^2$. Note that \[3(ab + bc + ca) \le \dfrac{9abc}{a + b + c} + 2(a^2 + b^2 + c^2) \iff 6u^3 -7uv^2 + v^3 \ge 0.\]Let $t = \dfrac{u}{v}$, dividing the above relation by $v^3$ we want to show \[ 6t^3 - 7t + 1 \ge 0 \iff (t - 1)(6t^2 + 6t - 1) \ge 0 \]which is true since $t \ge 1$. Now note that $\dfrac{9}{a + b + c} \le 1$ by Cauchy-Schwarz so adding this to what we proved yields the original inequality.
eg4334
20.11.2024 06:39
It is apparent that $\frac{9}{a+b+c} \leq 1$, so it remains to show that $$3(ab+bc+ac)(a+b+c) \leq 9abc+2(a^2+b^2+c^2)(a+b+c)$$$$\sum a^2b \leq 2 \sum a^3$$which is apparent by Muirheads.