Nonsymmetric inequality for sum of three fractions

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Nonsymmetric inequality for sum of three fractions

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Source: Macedonian TST for IMO 2013 - P1 day 2

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#1 h Mar 28, 2021, 11:02 PM

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Let $a>0,b>0,c>0$ and $a+b+c=1$ . Show the inequality
$\frac{a^4+b^4}{a^2+b^2}+\frac{b^3+c^3}{b+c} + \frac{2a^2+b^2+2c^2}{2} \geq \frac{1}{2}$

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#2 h Mar 29, 2021, 12:14 AM • 1 Y

Y by steppewolf

By Cauchy, $(a^4+b^4)\ge (a^2+b^2)^2/2$ . Hence, the given object is lower bounded by
$\frac{3a^2+2b^2+2c^2 +2b^2-2bc+2c^2}{2} = \frac{3a^2+4b^2+4c^2 -2bc}{2}.$ Now, using (a) $b^2+c^2-2bc\ge 0$ ; and (b) $3a^2+3b^2+3c^2\ge (a+b+c)^2=1$ valid by the Cauchy-Schwarz, we complete the proof.

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sqing

#3 h Mar 29, 2021, 6:13 AM • 1 Y

Y by steppewolf

steppewolf wrote:

Let $a>0,b>0,c>0$ and $a+b+c=1$ . Show the inequality
$\frac{a^4+b^4}{a^2+b^2}+\frac{b^3+c^3}{b+c} + \frac{2a^2+b^2+2c^2}{2} \geq \frac{1}{2}$

https://artofproblemsolving.com/community/c6h1195489p5852461
Let $a>0,b>0,c>0$ and $a+b+c=1$ . Show the inequality
$\frac{1}{2}\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)^2\geq\frac{a^4+b^4}{a^2+b^2}+\frac{b^3+c^3}{b+c}+\frac{2a^2+b^2+2c^2}{2} \geq \frac{1}{2}$

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#4 h Mar 29, 2021, 8:19 AM

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Muirhead on LHS first and second term gives $a^4 + b^4 \ge ab(a^2 + b^2)$ and $b^3 + c^3 \ge bc(b + c)$ . Then, AM-GM gives $\frac{a^2 + c^2}{2} \ge ac$ and C-S gives $\frac{a^2 + b^2 + c^2}{2} \ge \frac{1}{6}$ .

Thus, $LHS \ge ab + bc + bc + \dfrac{1}{6}$ and since $ab + bc + ac \ge \dfrac{1}{3}$ (which is trivial), $LHS \ge \dfrac{1}{2}$ . $\blacksquare$

This post has been edited 1 time. Last edited by CadmiumDwight, Mar 29, 2021, 8:20 AM

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#5 h Mar 29, 2021, 11:08 AM

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CadmiumDwight wrote:

$ab + bc + ac \ge \dfrac{1}{3}$ (which is trivial)

Are you sure?

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steppewolf wrote:

Let $a>0,b>0,c>0$ and $a+b+c=1$ . Show the inequality
$\frac{a^4+b^4}{a^2+b^2}+\frac{b^3+c^3}{b+c} + \frac{2a^2+b^2+2c^2}{2} \geq \frac{1}{2}$

We have, by Chebyshev and AM-GM:

$LHS\ge \frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{2a^2+b^2+2c^2}{2}=\frac{3}{2}(a^2+b^2+c^2)\ge \frac{3}{2} \cdot \frac{(a+b+c)^2}{3}=\frac{1}{2}$

This post has been edited 2 times. Last edited by Math00954, Mar 29, 2021, 11:13 AM

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F10tothepowerof34

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F10tothepowerof34

#6 h Jun 3, 2023, 12:09 AM

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Notice that, $\frac{a^4+b^4}{a^2+b^2}\ge\frac{a^2+b^2}{2}\Longrightarrow 2a^4+2b^4\ge(a^2+b^2)^2=a^2+2a^2b^2+b^4\Longrightarrow a^4+b^4\overset{\text{AM-GM}}{\ge}2a^2b^2$
Furthermore, $\frac{b^3+c^3}{b+c}\ge\frac{b^2+c^2}{2b^3+2c^3}\Longrightarrow \ge(b^2+c^2)(b+c)=b^3+b^2c+c^2b+c^3\Longrightarrow b^3+c^3\ge b^2c+c^2b\Longrightarrow (b+c)(b^2+c^2-bc)\ge bc(b+c)=b^2c+c^2b$
Thus by summing the previously stated inequalities, $\sum\frac{a^4+b^4}{a^2+b^2}\ge\frac{3\sum_{cyc}a^2}{2}=\frac{3}{2}\left(\bigg(\sum_{cyc}a\bigg)^2-2\sum_{cyc}ab\right)=\frac{3}{2}\bigg(1-2\sum_{cyc}ab\bigg)$ , thus the inequality boils down to proving: $\frac{3}{2}-3\sum_{cyc}ab\ge\frac{1}{2}\Longleftrightarrow 1\ge3\sum_{cyc}ab$ thus after we homogenize the $LHS$ , the inequality turns into $\bigg(\sum_{cyc}a\bigg)^2\ge3\sum_{cyc}ab$ which is clearly true $\blacksquare$ .

This post has been edited 2 times. Last edited by F10tothepowerof34, Jun 19, 2023, 3:13 AM

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