Given the inequalities: $a)$ $\left(\frac{2a}{b+c}\right)^2+\left(\frac{2b}{a+c}\right)^2+\left(\frac{2c}{a+b}\right)^2\geq \frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ $b)$ $\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+9$ For each of them either prove that it holds for all positive real numbers $a$, $b$, $c$ or present a counterexample $(a,b,c)$ which doesn't satisfy the inequality.
Problem
Source: 2022 Bulgarian Spring Math Competition, Problem 8.3
Tags: algebra, Inequality, inequalities
sqing
27.03.2022 12:31
Let $a,b,c$ be positive real numbers. Prove that$$\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+9$$
Marinchoo wrote:
Given the inequalities:
$a)$ $\left(\frac{2a}{b+c}\right)^2+\left(\frac{2b}{a+c}\right)^2+\left(\frac{2c}{a+b}\right)^2\geq \frac{a}{c}+\frac{b}{a}+\frac{c}{b}$
$b)$ $\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+9$
For each of them either prove that it holds for all positive real numbers $a$, $b$, $c$ or present a counterexample $(a,b,c)$ which doesn't satisfy the inequality.
sqing
27.03.2022 12:52
Let $a,b,c$ be positive real numbers. Prove that$$\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq 4\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)$$
TYT
27.03.2022 13:03
Marinchoo wrote: Given the inequalities: $a)$ $\left(\frac{2a}{b+c}\right)^2+\left(\frac{2b}{a+c}\right)^2+\left(\frac{2c}{a+b}\right)^2\geq \frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ $b)$ $\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+9$ For each of them either prove that it holds for all positive real numbers $a$, $b$, $c$ or present a counterexample $(a,b,c)$ which doesn't satisfy the inequality. The inequality $a)$ is false for $a = 0^+, b = c = 1$. As for b), using the inequality $x^2 + y^2 + z^2 \geq xy + yz + zx$ we have \begin{align*} \sum \left( \frac{a + b}{c} \right)^2 &\geq \sum \frac{(a + b)(b + c)}{ca} = \sum \frac{ab + ca + b^2 + bc}{ca} \\ &= \sum \frac{b}{c} + 3 + \sum \frac{b^2}{ca} + \sum \frac{b}{a} \\ &\geq \sum \frac{a}{b} + 3 + 3 + 3 \\ &= \sum \frac{a}{b} + 9 \end{align*}
mihaig
27.03.2022 13:08
The second is true and nice.
sqing
27.03.2022 13:18
Let $a,b,c$ be positive real numbers. Prove that$$\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+9$$https://artofproblemsolving.com/community/c6h486634p5449967
TYT wrote:
Marinchoo wrote:
Given the inequalities:
$a)$ $\left(\frac{2a}{b+c}\right)^2+\left(\frac{2b}{a+c}\right)^2+\left(\frac{2c}{a+b}\right)^2\geq \frac{a}{c}+\frac{b}{a}+\frac{c}{b}$
$b)$ $\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+9$
For each of them either prove that it holds for all positive real numbers $a$, $b$, $c$ or present a counterexample $(a,b,c)$ which doesn't satisfy the inequality.
The inequality $a)$ is false for $a = 0^+, b = c = 1$. As for b), using the inequality $x^2 + y^2 + z^2 \geq xy + yz + zx$ we have
\begin{align*}
\sum \left( \frac{a + b}{c} \right)^2 &\geq \sum \frac{(a + b)(b + c)}{ca} = \sum \frac{ab + ca + b^2 + bc}{ca} \\
&= \sum \frac{b}{c} + 3 + \sum \frac{b^2}{ca} + \sum \frac{b}{a} \\
&\geq \sum \frac{a}{b} + 3 + 3 + 3 \\
&= \sum \frac{a}{b} + 9
\end{align*}
VicKmath7
27.03.2022 14:14
It is equivalent to $\sum \frac {a^2}{b^2}+\sum \frac {2ab} {c^2} \geq \sum \frac {a}{b} +9$ (the first two sums are symmetrical, the third is cyclic). Now sum the bounds $\frac {a^2}{b^2} \geq 2\frac {a}{b}-1$ cyclically (not symmetrically!) and apply AM-GM for all 12 terms you obtain at the end and we're done.
mihaig
27.03.2022 14:17
TYT wrote: Marinchoo wrote: Given the inequalities: $a)$ $\left(\frac{2a}{b+c}\right)^2+\left(\frac{2b}{a+c}\right)^2+\left(\frac{2c}{a+b}\right)^2\geq \frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ $b)$ $\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+9$ For each of them either prove that it holds for all positive real numbers $a$, $b$, $c$ or present a counterexample $(a,b,c)$ which doesn't satisfy the inequality. The inequality $a)$ is false for $a = 0^+, b = c = 1$. As for b), using the inequality $x^2 + y^2 + z^2 \geq xy + yz + zx$ we have \begin{align*} \sum \left( \frac{a + b}{c} \right)^2 &\geq \sum \frac{(a + b)(b + c)}{ca} = \sum \frac{ab + ca + b^2 + bc}{ca} \\ &= \sum \frac{b}{c} + 3 + \sum \frac{b^2}{ca} + \sum \frac{b}{a} \\ &\geq \sum \frac{a}{b} + 3 + 3 + 3 \\ &= \sum \frac{a}{b} + 9 \end{align*} Bravo
sqing
27.03.2022 14:24
TYT wrote:
Marinchoo wrote:
Given the inequalities:
$a)$ $\left(\frac{2a}{b+c}\right)^2+\left(\frac{2b}{a+c}\right)^2+\left(\frac{2c}{a+b}\right)^2\geq \frac{a}{c}+\frac{b}{a}+\frac{c}{b}$
$b)$ $\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+9$
For each of them either prove that it holds for all positive real numbers $a$, $b$, $c$ or present a counterexample $(a,b,c)$ which doesn't satisfy the inequality.
The inequality $a)$ is false for $a = 0^+, b = c = 1$. As for b), using the inequality $x^2 + y^2 + z^2 \geq xy + yz + zx$ we have
\begin{align*}
\sum \left( \frac{a + b}{c} \right)^2 &\geq \sum \frac{(a + b)(b + c)}{ca} = \sum \frac{ab + ca + b^2 + bc}{ca} \\
&= \sum \frac{b}{c} + 3 + \sum \frac{b^2}{ca} + \sum \frac{b}{a} \\
&\geq \sum \frac{a}{b} + 3 + 3 + 3 \\
&= \sum \frac{a}{b} + 9
\end{align*}
Let $a,b,c$ be positive real numbers. Prove that$$\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} +6$$
Marinchoo
27.03.2022 15:31
\[\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2=\frac{1}{2}\sum\limits_{cyc} \left(\frac{a^2}{c^2}+\frac{c^2}{b^2}\right)+\left[\sum\limits_{cyc} \frac{b^2}{c^2}+2\sum\limits_{cyc}\frac{ab}{c^2}\right]\overset{AMGM}{\geq} \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+9\]
Marinchoo
27.03.2022 15:39
sqing wrote: Let $a,b,c$ be positive real numbers. Prove that$$\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} +6$$ \[\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2=\frac{1}{2}\sum\limits_{cyc} \left(\frac{a^2}{c^2}+\frac{c^2}{b^2}\right)+\frac{1}{2}\sum\limits_{cyc} \left(\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)+2\sum\limits_{cyc}\frac{ab}{c^2}\overset{AMGM}{\geq} \sum\limits_{cyc} \frac{a}{b}+\sum\limits_{cyc} \frac{b}{a}+6\]
sqing
27.03.2022 17:46
sqing wrote: Let $a,b,c$ be positive real numbers. Prove that$$\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq 4\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)$$ Solution of Zhangyanzong: By AM-GM, $$\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq 4\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)-12$$$$=4\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)+4\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}-3\right)\geq4\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)$$
sqing
27.03.2022 18:47
Let $a,b,c$ be positive real numbers. Prove that$$\left(\frac{b+c}{a}\right)^2+\frac{2(a+b)(c+a)}{bc}\geq \frac{4b}{a}+8\sqrt{\frac{a}{b}} $$
sqing
28.03.2022 10:01
Marinchoo wrote:
sqing wrote:
Let $a,b,c$ be positive real numbers. Prove that$$\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} +6$$
\[\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2=\frac{1}{2}\sum\limits_{cyc} \left(\frac{a^2}{c^2}+\frac{c^2}{b^2}\right)+\frac{1}{2}\sum\limits_{cyc} \left(\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)+2\sum\limits_{cyc}\frac{ab}{c^2}\overset{AMGM}{\geq} \sum\limits_{cyc} \frac{a}{b}+\sum\limits_{cyc} \frac{b}{a}+6\]
Attachments:
sqing
28.03.2022 15:35
sqing wrote: Let $a,b,c$ be positive real numbers. Prove that$$\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq 4\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)$$
Attachments:
sqing
01.04.2022 09:15
sqing wrote: Let $a,b,c$ be positive real numbers. Prove that$$\left(\frac{b+c}{a}\right)^2+\frac{2(a+b)(c+a)}{bc}\geq \frac{4b}{a}+8\sqrt{\frac{a}{b}} $$
Attachments:
