Inequality with 3 variables and a special condition

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Inequality with 3 variables and a special condition

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Source: Azerbaijan Al-Khwarizmi IJMO TST 2024

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#1 h Apr 29, 2025, 5:06 PM • 1 Y

Y by TDVOLIMPTEAM

For positive real numbers $a,b,c$ we have $3abc \geq ab+bc+ca$ .
Prove that:

$\frac{1}{a^3+b^3+c}+\frac{1}{b^3+c^3+a}+\frac{1}{c^3+a^3+b} \leq \frac{3}{a+b+c}$ .

Determine the equality case.

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arqady

#2 h Apr 29, 2025, 5:29 PM • 3 Y

Y by MuradSafarli, Nuran2010, FarrukhBurzu

Nuran2010 wrote:

For positive real numbers $a,b,c$ we have $3abc \geq ab+bc+ca$ .
Prove that:

$\frac{1}{a^3+b^3+c}+\frac{1}{b^3+c^3+a}+\frac{1}{c^3+a^3+b} \leq \frac{3}{a+b+c}$ .

Determine the equality case.

Use C-S and AM-GM.

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#3 h Apr 29, 2025, 10:53 PM • 1 Y

Y by Nuran2010

Reduce to a common denominator via Cauchy-Schwarz exercise (similar to problems from JBMO 2019 and 2023 shortlists). We have $(a^3 + b^3 + c)\left(\frac{1}{a} + \frac{1}{b} + c\right) \geq (a+b+c)^2$ and the analogous ones, so the original left-hand side does not exceed
$\frac{2\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) + a+b+c}{(a+b+c)^2}$ and we wish the latter to not exceed $\frac{3}{a+b+c}$ . Equivalently, we want $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq a+b+c$ . But actually, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq 3$ from the problem condition, while $3abc \geq ab + bc + ca \geq 3\sqrt[3]{a^2b^2c^2}$ by AM-GM implies $abc \geq 1$ , thus $a+b+c \geq 3\sqrt[3]{abc} \geq 3$ by AM-GM and we are done. The AM-GM-s and $abc \geq 1$ require $a=b=c=1$ for equality and substituting in the initial inequality indeed shows that it is an equality in this case.

This post has been edited 1 time. Last edited by Assassino9931, Apr 29, 2025, 10:54 PM

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sqing

#4 h May 1, 2025, 8:58 AM

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Prove that for all non-negative real numbers $x,y,z$ , not all equal to $0$ , the following inequality holds
$\displaystyle \dfrac{2x^2-x+y+z}{x+y^2+z^2}+\dfrac{2y^2+x-y+z}{x^2+y+z^2}+\dfrac{2z^2+x+y-z}{x^2+y^2+z}\geq 3$ Let $a, b, c$ be positive real numbers such that $abc = \frac {2} {3}.$ Prove that:
$\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} \geqslant \frac {a+b+c} {a^3+b ^ 3 + c ^ 3}.$ (JBMO 2019 and 2023 shortlists)

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#5 h May 2, 2025, 12:56 AM • 1 Y

Y by ehuseyinyigit

@Sqing No, I meant A5 in 2019 and A4 in 2023

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sqing

#6 h May 2, 2025, 1:26 AM • 1 Y

Y by ehuseyinyigit

Assassino9931 wrote:

@Sqing No, I meant A5 in 2019 and A4 in 2023

Let $a, b, c, d$ be positive real numbers such that $abcd = 1$ . Prove the inequality
$\frac{1}{a^3 + b + c + d} +\frac{1}{a + b^3 + c + d}+\frac{1}{a + b + c^3 + d} +\frac{1}{a + b + c + d^3} \leq \frac{a+b+c+d}{4}$
Let $a,b,c,d$ be positive real numbers with $abcd=1$ . Prove that
$\sqrt{\frac{a}{b+c+d^2+a^3}}+\sqrt{\frac{b}{c+d+a^2+b^3}}+\sqrt{\frac{c}{d+a+b^2+c^3}}+\sqrt{\frac{d}{a+b+c^2+d^3}} \leq 2$

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mihaig

#7 h May 2, 2025, 9:20 AM

Y by

arqady wrote:

Nuran2010 wrote:

For positive real numbers $a,b,c$ we have $3abc \geq ab+bc+ca$ .
Prove that:

$\frac{1}{a^3+b^3+c}+\frac{1}{b^3+c^3+a}+\frac{1}{c^3+a^3+b} \leq \frac{3}{a+b+c}$ .

Determine the equality case.

Use C-S and AM-GM.

Beautiful

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sqing

#8 h May 2, 2025, 9:48 AM

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Let $a,b>0$ and $a^3+1= a^2(b+1).$ Prove that
$\frac{1}{a^3+b^3+1}+\frac{1}{a^3+b+1}+\frac{1}{b^3+a+1} \leq \frac{3}{ a+b+1}$

This post has been edited 1 time. Last edited by sqing, May 2, 2025, 10:06 AM

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sqing

#9 h May 2, 2025, 10:05 AM

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Let $a,b>0$ and $a^2+b^2+ab=\frac{3}{4}.$ Prove that
$\frac{1}{a^3+b^3+\frac{1}{2}}+\frac{1}{a^3+b+\frac{1}{8}}+\frac{1}{b^3+a+\frac{1}{8}} \leq \frac{6}{ a+b+\frac{1}{2}}$

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