Inequality with one variable rational functions

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Inequality with one variable rational functions

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Source: 2022 Junior Macedonian Mathematical Olympiad P2

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7 posts

#1 h Jun 7, 2022, 9:05 PM

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Let $a$ , $b$ and $c$ be positive real numbers such that $a+b+c=3$ . Prove the inequality
$\frac{a^3}{a^2+1}+\frac{b^3}{b^2+1}+\frac{c^3}{c^2+1} \geq \frac{3}{2}.$
Proposed by Anastasija Trajanova

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#2 h Jun 7, 2022, 9:19 PM • 2 Y

Y by teomihai, MihaiT

$\frac{a^3}{a^2+1} \geq \frac{2a-1}{2}$
So $\frac{a^3}{a^2+1}+\frac{b^3}{b^2+1}+\frac{c^3}{c^2+1} \geq \frac{2(a+b+c)-3}{2}=\frac{3}{2}$

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sqing

#3 h Jun 8, 2022, 12:53 AM • 1 Y

Y by Mango247

Let $a$ , $b$ and $c$ be positive real numbers such that $a+b+c=3$ . Prove the inequality
$\frac{a^3}{a^2+b}+\frac{b^3}{b^2+c}+\frac{c^3}{c^2+a} \geq\frac{3}{2}\geq \frac{a^2}{a^3+1}+\frac{b^2}{b^3+1}+\frac{c^2}{c^3+1}$ $\frac{a^3}{a^2+b+c}+\frac{b^3}{b^2+c+a}+\frac{c^3}{c^2+a+b} \geq 1$ $\frac{a^3}{a^2+1}+\frac{b^3}{b^2+1}+\frac{c^3}{c^2+1} \leq\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$ h

This post has been edited 2 times. Last edited by sqing, Jun 8, 2022, 1:17 AM

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#4 h Jun 8, 2022, 1:38 AM

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Just note that $\frac{a^3}{a^2+1} = a\cdot \left(1-\frac{1}{a^2+1}\right)$ , and thus it boils down proving
$\frac32 \ge \sum \frac{a}{a^2+1}$ . But as $a/(a^2+1)\le 1/2$ we are done.

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sqing

#5 h Jun 14, 2022, 2:43 AM • 1 Y

Y by Mango247

Let $a$ , $b$ and $c$ be non-negative real numbers such that $a+b+c=3$ . Prove that
$\frac{27}{10}\geq\frac{a^3}{a^2+1}+\frac{b^3}{b^2+1}+\frac{c^3}{c^2+1} \geq \frac{3}{2}$ $\frac{3}{2}\geq \frac{a^2}{a^3+1}+\frac{b^2}{b^3+1}+\frac{c^2}{c^3+1} \geq\frac{9}{28}$

This post has been edited 2 times. Last edited by sqing, Jun 14, 2022, 2:59 AM

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AlanLG

#6 h Jun 22, 2022, 11:49 PM

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It is not difficult to prove that

$\frac{x^3}{x^2+1}\geq \frac{1}{2}+x-1$
holds for all $x$ , summing the result inequalities we are done.

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#7 h Sep 3, 2022, 5:06 AM

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FALSEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

This post has been edited 1 time. Last edited by strong_boy, Sep 3, 2022, 7:13 AM

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#8 h Sep 3, 2022, 5:33 AM

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strong_boy wrote:

Cute problem :
We know $a^2 +1 \geq 2a$ . (

)
By (

) and original problem we can see :
$\frac{a^3}{a^2+1}+\frac{b^3}{b^2+1}+\frac{c^3}{c^2+1} \geq \frac{a^3}{2a}+\frac{a^3}{2a}+\frac{a^3}{2a} = a^2+b^2+c^2$
Now we only need prove $a^2+b^2+c^2 \geq 3$ . and it is true beacuase $3(a^2+b^2+c^2) \geq (a+b+c)^2$ . $\blacksquare$

$X \ge Y$ doesn't mean $\frac{Z}{X} \ge \frac{Z}{Y}$ , this is false.

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#9 h Jun 23, 2023, 12:02 PM • 1 Y

Y by teomihai

Tangent line method...

$\frac{x^3}{x^2+1}\geq x-\frac{1}{2}\Leftrightarrow \frac{1}{2}(x-1)^2\geq0$
$\sum_{cyc}\frac{a^3}{a^2+1}\geq\sum_{cyc}(a-\frac{1}{2})=\frac{3}{2}$

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#10 h Aug 14, 2024, 11:33 PM

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Write $\frac{a^3}{a^2+1}$ as $a-\frac{a}{a^2+1}$ . Summing, we just want to prove $3-\sum_{cyc}\frac{a}{a^2+1}\geq \frac{3}{2}$ or rearranging, $\sum_{cyc}\frac{a}{a^2+1}\leq \frac{3}{2}$ . We claim $\frac{a}{a^2+1}\leq \frac{1}{2}$ . This is trivial as upon rearranging, $(a-1)^2\geq 0$ . Summing finishes.

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43156 posts

sqing

#11 h Aug 15, 2024, 2:16 AM

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Let $a,b,c >0$ and $a+b+c=3$ . Prove that
$\frac{a^3}{a^2+2}+\frac{ b^3}{b^2+8}+\frac{c^3}{c^2+2} \geq \frac{27}{41}$ $\frac{a^3}{a^2+2}+\frac{ b^3}{b^2+3}+\frac{c^3}{c^2+2} \geq \frac{27(5-\sqrt 6)}{76}$

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#12 h Aug 17, 2024, 2:08 PM

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sqing wrote:

Let $a$ , $b$ and $c$ be non-negative real numbers such that $a+b+c=3$ . Prove that
$\frac{27}{10}\geq\frac{a^3}{a^2+1}+\frac{b^3}{b^2+1}+\frac{c^3}{c^2+1} \geq \frac{3}{2}$ $\frac{3}{2}\geq \frac{a^2}{a^3+1}+\frac{b^2}{b^3+1}+\frac{c^2}{c^3+1} \geq\frac{9}{28}$

The left hand side of the second inequality is due to tangent line trick. RTP $f\left(x\right)-f\left(1\right)-\left(x-1\right)f'\left(1\right)\leq 0$ if $f\left(x\right)=\frac{x^{2}}{x^{3}+1}$ . Rearrange as showing $x^{4}+x^{3}-4x^{2}+x+1\geq 0$ . Standard calc exercise. Derivative tells us local minimum at $x=1$ , other minimums are at negative $x$ values. Easy to check that this gives $1+1-4+1+1=0$ as the $y$ coordinate hence everything else is just nonnegative.

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3 posts

#13 h Mar 9, 2025, 12:53 PM

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from AM-GM we have:
a^2+1>=2a,
b^2+1>=2b,
c^2+1>=2c
Hence (a^3)/2a + (b^3)/2b + (c^3)/2c >= 3/2 <=>
(a^2)/2 + (b^2)/2 + (c^2)/2 >= 3/2
from the special case of the B-C-S(andreescu) inequality we have:
[(a+b+c)^2] / (2+2+2) >= 3/2 <=>
(3^2) / 6 >= 3/2 <=>
3/2 >= 3/2 which applies

This post has been edited 1 time. Last edited by basilis, Mar 11, 2025, 10:59 AM

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#15 h Mar 10, 2025, 4:08 PM • 1 Y

Y by basilis

$LHS = \sum\limits_{\text{cyc}} \frac{a^3 + a - a}{a^2 + 1}$
= $\sum\limits_{\text{cyc}} a - \frac{a}{a^2+1}$
$\ge \sum\limits_{\text{cyc}} a - \frac{a}{2a}$ (from $AM-GM$ )
= $RHS$ $\blacksquare$ .

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IEatProblemsForBreakfast

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IEatProblemsForBreakfast

#16 h May 22, 2025, 7:30 PM

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This post has been edited 4 times. Last edited by IEatProblemsForBreakfast, Jun 17, 2025, 5:48 PM
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