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sqing

#1 h Mar 28, 2025, 3:10 PM

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Let $a,b,c\geq 0$ and $a+b+c=3.$ Provethat
$\frac{3}{11}\leq \frac{a}{a^2+2}+\frac{b}{b^2+2}+\frac{c}{c^2+2}\leq1$

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sqing

#2 h Mar 28, 2025, 3:25 PM

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Let $a,b\geq 0$ and $a^2+b^2+a+b =4.$ Provethat
$\frac{1+3\sqrt{17}}{38} \leq \frac{a}{a^2+2}+\frac{b}{b^2+2} \leq \frac{2}{3}$ Let $a,b\geq 0$ and $a^2+b^2+a+b =6.$ Provethat
$\frac{1}{3}\leq \frac{a}{a^2+2}+\frac{b}{b^2+2} \leq \frac{1+5\sqrt{13}}{27}$ Let $a,b\geq 0$ and $a^2+b^2+a+b+ab =5.$ Provethat
$\frac{3+7\sqrt{21}}{102} \leq \frac{a}{a^2+2}+\frac{b}{b^2+2} \leq \frac{2}{3}$

This post has been edited 1 time. Last edited by sqing, Mar 28, 2025, 3:35 PM

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Primeniyazidayi

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Primeniyazidayi

#3 h Mar 28, 2025, 3:27 PM

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For the right side:Apply AM-GM to a^2+2, subtract both sides with 3 and finish with Titu's Lemma.

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#4 h Mar 28, 2025, 3:29 PM

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sqing wrote:

Let $a,b,c\geq 0$ and $a+b+c=3.$ Provethat
$\frac{3}{11}\leq \frac{a}{a^2+2}+\frac{b}{b^2+2}+\frac{c}{c^2+2}\leq1$

sketch

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Primeniyazidayi

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Primeniyazidayi

#5 h Mar 28, 2025, 3:48 PM

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For the left side:Apply Titu's Lemma

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imnotgoodatmathsorry

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imnotgoodatmathsorry

#6 h Mar 28, 2025, 4:00 PM

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sqing wrote:

Let $a,b,c\geq 0$ and $a+b+c=3.$ Provethat
$\frac{3}{11}\leq \frac{a}{a^2+2}+\frac{b}{b^2+2}+\frac{c}{c^2+2}\leq1$

Solution 1:
We have: $\frac{a}{a^2+2} - [\frac{1}{3} + \frac{1}{9}(a-1)] = \frac{-(a-1)^2(a+4)}{9(a^2+2)} \le 0$ so $\frac{a}{a^2+2} \le \frac{1}{3} + \frac{1}{9}(a-1)$
$\rightarrow LHS = \sum{\frac{a}{a^2+2}} \le \sum{\frac{1}{3} + \frac{1}{9}(a-1)} = 1$ cuz $a+b+c=3$
The equality occurs when $(a;b;c)=(1;1;1)$
Solution 2:
By $AM-GM$ we have $a^2 +2 \ge 2a+1$ so:
$\rightarrow LHS = \sum{\frac{a}{a^2+2}} \le \sum{\frac{a}{2a+1}} = \frac{3}{2} - \sum{\frac{1}{2(2a+1)}}$
so now we need to prove $\sum{\frac{1}{2a+1}} \ge 1$ which is trival by Titu.

This post has been edited 3 times. Last edited by imnotgoodatmathsorry, Mar 28, 2025, 4:12 PM
Reason: Wrong number

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sqing

#8 h Mar 29, 2025, 2:49 AM

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Thank you.

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#9 h Mar 29, 2025, 3:15 AM

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sqing wrote:

Let $a,b,c\geq 0$ and $a+b+c=3.$ Provethat
$\frac{3}{11}\leq \frac{a}{a^2+2}+\frac{b}{b^2+2}+\frac{c}{c^2+2}\leq1$

Note that the inequality $\begin{align*} \frac{x}{11} \leq \frac{x}{x^2 + 2} \leq \frac{x}{9} + \frac{2}{9} \end{align*}$
Holds over the interval $[0, 3]$ . Cyclically summing this inequality yields the desired result.

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