# Problem 297 Van Khea

If $a, b, c$ are positive real numbers. Prove that:
$\displaystyle \frac{a+b}{b+c}.a^2+\frac{b+c}{c+a}.b^2+\frac{c+a}{a+b}.c^2\geq \frac{1}{3}(a+b+c)^2$
Proof
From $Cauchy-Schwarz$ we have
$\displaystyle \frac{a+b}{b+c}.a^2+\frac{b+c}{c+a}.b^2+\frac{c+a}{a+b}.c^2\geq \frac{(a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a})^2}{2(ab+bc+ca)}$
But for $a, b, c$ are positive real numbers then $3(ab+bc+ca)\leq (a+b+c)^2$
$\displaystyle \Rightarrow \frac{a+b}{b+c}.a^2+\frac{b+c}{c+a}.b^2+\frac{c+a}{a+b}.c^2\geq \frac{3(a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a})^2}{2(a+b+c)^2}$
Thus, we need to prove that
$\displaystyle a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a}\geq \frac{\sqrt{2}}{3}(a+b+c)^2$
We have $\displaystyle a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a}$$\displaystyle =a^2\sqrt{1+\frac{b}{a}}+b^2\sqrt{1+\frac{c}{b}}+c^2\sqrt{1+\frac{a}{c}}$$\displaystyle =\frac{a^2}{\sqrt{\frac{a}{a+b}}}+\frac{b^2}{\sqrt{\frac{b}{b+c}}}+\frac{c^2}{\sqrt{\frac{c}{c+a}}}$$\displaystyle \geq \frac{(a+b+c)^2}{\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}}$
From $Problem$ $Vasile Cirtoaje$ we have
$\displaystyle \sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\leq \frac{3}{\sqrt{2}}$
$\displaystyle \Rightarrow a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a}\geq \frac{\sqrt{2}}{3}(a+b+c)^2$
$\displaystyle \Rightarrow (a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a})^2\geq \frac{2}{9}(a+b+c)^4$
$\displaystyle \Rightarrow \frac{a+b}{b+c}.a^2+\frac{b+c}{c+a}.b^2+\frac{c+a}{a+b}.c^2\geq \frac{1}{3}(a+b+c)^2$
Therefore the proof is completed. Equality occurs for $a=b=c$
របៀបទីពីរ ( Rainymathboy)
សន្មត $a\ge b\ge c$
ទាញបាន $\displaystyle {{a}^{2}}\ge {{b}^{2}}\ge {{c}^{2}}$
$\displaystyle a+b\ge c+a\ge b+c$ និង​ $\frac{1}{b+c}\ge \frac{1}{c+a}\ge \frac{1}{a+b}$
នោះ $\displaystyle \frac{{{a}^{2}}}{b+c}\ge \frac{{{b}^{2}}}{c+a}\ge \frac{{{c}^{2}}}{a+b}$
តាមវិសមភាព​តំរៀប (Rearrangement​ inequality) យើងមានៈ
បើ​ $\displaystyle {{a}_{1}}\ge {{a}_{2}}\ge \ldots \ge {{a}_{n}}$​ និង ${{b}_{1}}\ge {{b}_{2}}\ge \ldots \ge {{b}_{n}}$
តាង $\displaystyle A={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+\ldots +{{a}_{n}}{{b}_{n}}$
$\displaystyle B={{a}_{1}}{{b}_{\sigma (1)}}+{{a}_{2}}{{b}_{\sigma (2)}}+\ldots +{{a}_{n}}{{b}_{\sigma (n)}}$
$\displaystyle C={{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+\ldots +{{a}_{n}}{{b}_{1}}$
នាំអោយ $A\ge B\ge C$
យើងទាញបានៈ
$\displaystyle A=\frac{{{a}^{2}}}{b+c}(a+b)+\frac{{{b}^{2}}}{c+a}(c+a)+\frac{{{c}^{2}}}{a+b}(b+c)$
$\displaystyle B=\frac{{{a}^{2}}}{b+c}(a+b)+\frac{{{b}^{2}}}{c+a}(b+c)+\frac{{{c}^{2}}}{a+b}(c+a)$
$\displaystyle C=\frac{{{a}^{2}}}{b+c}(b+c)+\frac{{{b}^{2}}}{c+a}(c+a)+\frac{{{c}^{2}}}{a+b}(a+b)$
យើងយកលក្ខខណ្ឌ​ $B\ge C$ មក​ប្រើ​នោះយើងទាញបានៈ
$\displaystyle \frac{{{a}^{2}}}{b+c}(a+b)+\frac{{{b}^{2}}}{c+a}(b+c)+\frac{{{c}^{2}}}{a+b}(c+a)$
$\displaystyle \ge \frac{{{a}^{2}}}{b+c}(b+c)+\frac{{{b}^{2}}}{c+a}(c+a)+\frac{{{c}^{2}}}{a+b}(a+b)={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$
$\displaystyle a^2+b^2+c^2\geq \frac{1}{3}(a+b+c)^2$
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល $a=b=c$

### 17 Responses to Problem 297 Van Khea

1. ដោយ​មិន​បាត់​បង់​លក្ខណៈ​ទូទៅ (W.L.O.G) ខ្ញុំ​សន្មត​ថា $a\ge b\ge c$
យើង​បាន $\left( a+b \right){{a}^{2}}\ge \left( b+c \right){{b}^{2}}\ge \left( c+a \right){{c}^{2}}$
និង $\frac{1}{c+a}\ge \frac{1}{b+a}\ge \frac{1}{a+b}$
តាម​វិសមភាព​តម្រៀប (Rearrangement) យើង​បាន
$\frac{a+b}{b+c}.{{a}^{2}}+\frac{b+c}{c+a}.{{b}^{2}}+\frac{c+a}{a+b}.{{c}^{2}}\ge \frac{a+b}{a+b}.{{a}^{2}}+\frac{b+c}{b+c}.{{b}^{2}}+\frac{c+a}{c+a}.{{c}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$
តែ​តាម​ Lemma Cauchy-Schwarz
${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=\frac{{{a}^{2}}}{1}+\frac{{{b}^{2}}}{1}+\frac{{{c}^{2}}}{1}\ge \frac{{{\left( a+b+c \right)}^{2}}}{3}$

ដូចនេះ
$\frac{a+b}{b+c}.{{a}^{2}}+\frac{b+c}{c+a}.{{b}^{2}}+\frac{c+a}{a+b}.{{c}^{2}}\ge \frac{1}{3}{{(a+b+c)}^{2}}$

2. ហេតុ​អី​ខ្ញុំ​មើល​អត់​ឃើញ??
សាក​ល្បង​សារ​ជា​ថ្មី!
ដោយ​មិន​បាត់​បង់​លក្ខណៈ​ទូទៅ (W.L.O.G) ខ្ញុំ​សន្មត​ថា $a\ge b\ge c$
យើង​បាន $\left( a+b \right){{a}^{2}}\ge \left( b+c \right){{b}^{2}}\ge \left( c+a \right){{c}^{2}}$
និង $\frac{1}{c+a}\ge \frac{1}{b+a}\ge \frac{1}{a+b}$
តាម​វិសមភាព​តម្រៀប (Rearrangement) យើង​បាន
$\frac{a+b}{b+c}.{{a}^{2}}+\frac{b+c}{c+a}.{{b}^{2}}+\frac{c+a}{a+b}.{{c}^{2}}\ge \frac{a+b}{a+b}.{{a}^{2}}+\frac{b+c}{b+c}.{{b}^{2}}+\frac{c+a}{c+a}.{{c}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$
តែ​តាម​ Lemma Cauchy-Schwarz
${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=\frac{{{a}^{2}}}{1}+\frac{{{b}^{2}}}{1}+\frac{{{c}^{2}}}{1}\ge \frac{{{\left( a+b+c \right)}^{2}}}{3}$

ដូចនេះ
$\frac{a+b}{b+c}.{{a}^{2}}+\frac{b+c}{c+a}.{{b}^{2}}+\frac{c+a}{a+b}.{{c}^{2}}\ge \frac{1}{3}{{(a+b+c)}^{2}}$

3. vankhea says:

$(b+c)b^2\geq (c+a)c^2$ ?? សូមហេតុផល

4. rainymathboy says:

តាង​ខុស!
តាម​នេះ​វិញ​ប្រហែល​ជា​ត្រូវ​:
សន្មត $a\ge b\ge c$
ទាញបាន ${{a}^{2}}\ge {{b}^{2}}\ge {{c}^{2}}$
$a+b\ge c+a\ge b+c$ និង​ $\frac{1}{b+c}\ge \frac{1}{c+a}\ge \frac{1}{a+b}$
នោះ $\frac{{{a}^{2}}}{b+c}\ge \frac{{{b}^{2}}}{c+a}\ge \frac{{{c}^{2}}}{a+b}$
តាម Rearrangement ពេញ​លេញ:
$\frac{{{a}^{2}}}{b+c}(a+b)+\frac{{{b}^{2}}}{c+a}(c+a)+\frac{{{c}^{2}}}{a+b}(b+c)$
$\ge \frac{{{a}^{2}}}{b+c}(a+b)+\frac{{{b}^{2}}}{c+a}(b+c)+\frac{{{c}^{2}}}{a+b}(c+a)$
$\ge \frac{{{a}^{2}}}{b+c}(b+c)+\frac{{{b}^{2}}}{c+a}(c+a)+\frac{{{c}^{2}}}{a+b}(a+b)={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$
យក​អា​ពីរ​កង់​ចុង​ក្រោយ​មក​ប្រើ​ជា​ការ​ស្រេច

• rainymathboy says:

ត្រូវ​នៅ​លោក​បង? ខ្ញុំ​បុក​ពោះ​ខ្លាំង​ណាស់… 😀

• vankhea says:

នៅតែខុស
$\displaystyle \frac{a+b}{b+c}.a^2+\frac{b+c}{c+a}.b^2+\frac{c+a}{a+b}.c^2$$\displaystyle \neq \frac{a+b}{b+c}a^2+\frac{c+a}{c+a}b^2+\frac{b+c}{a+b}c^2$

• rainymathboy says:

ខ្ញុំ​សូម​វិចារនូវ​ វិសមភាព​Rearrangementបន្តិច:
បើ​ ${{a}_{1}}\ge {{a}_{2}}\ge \ldots \ge {{a}_{n}}$​ និង ${{b}_{1}}\ge {{b}_{2}}\ge \ldots \ge {{b}_{n}}$
តាង $A={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+\ldots +{{a}_{n}}{{b}_{n}}$
$B={{a}_{1}}{{b}_{\sigma (1)}}+{{a}_{2}}{{b}_{\sigma (2)}}+\ldots +{{a}_{n}}{{b}_{\sigma (n)}}$
$C={{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+\ldots +{{a}_{n}}{{b}_{1}}$
យើង​បាន $A\ge B\ge C$
តាម​ការ​តាង​របស់​ខ្ញុំ
$A=\frac{{{a}^{2}}}{b+c}(a+b)+\frac{{{b}^{2}}}{c+a}(c+a)+\frac{{{c}^{2}}}{a+b}(b+c)$
$B=\frac{{{a}^{2}}}{b+c}(a+b)+\frac{{{b}^{2}}}{c+a}(b+c)+\frac{{{c}^{2}}}{a+b}(c+a)$
$C=\frac{{{a}^{2}}}{b+c}(b+c)+\frac{{{b}^{2}}}{c+a}(c+a)+\frac{{{c}^{2}}}{a+b}(a+b)$
តែ​ខ្ញុំ​យក​តែ​លក្ខខណ្ឌ​ $B\ge C$ មក​ប្រើ​តែ​ប៉ុណ្ណោះ គឺ
$\frac{{{a}^{2}}}{b+c}(a+b)+\frac{{{b}^{2}}}{c+a}(b+c)+\frac{{{c}^{2}}}{a+b}(c+a)$
$\ge \frac{{{a}^{2}}}{b+c}(b+c)+\frac{{{b}^{2}}}{c+a}(c+a)+\frac{{{c}^{2}}}{a+b}(a+b)={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$

ត្រឹម​ត្រូវ​ទេ​បង?

5. vankhea says:

ខ្ញុំបានស្រាយទ្រឹស្ដីបទខាងលើរួចហើយ គឺវាពិតជានិច្ច។

ដូចនេះចំលើយ Rainy គឺត្រឹមត្រូវយកទទួលយកបាន អរគុណ Rainy ច្រើន។