1/(x ^ 2+y+z) +1/(x+y^2+z) + 1/(x+y+z^2) <=1, if x+y+z=3, x,y,z >=0

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1/(x ^ 2+y+z) +1/(x+y^2+z) + 1/(x+y+z^2) <=1, if x+y+z=3, x,y,z >=0

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Source: Tuymaada Junior 2006 p4

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#1 h May 12, 2019, 6:01 PM • 1 Y

Y by Adventure10

The sum of non-negative numbers $x$ , $y$ and $z$ is $3$ . Prove the inequality
${1 \over x ^ 2 + y + z} + {1 \over x + y ^ 2 + z} + {1 \over x + y + z ^ 2} \leq 1.$

This post has been edited 1 time. Last edited by parmenides51, May 12, 2019, 6:01 PM
Reason: latex

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#2 h May 12, 2019, 6:30 PM • 2 Y

Y by Adventure10, Mango247

perhaps, Jensen inequality?

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ytChen

#3 h May 12, 2019, 6:37 PM • 7 Y

Y by Pluto1708, White-Wolf, richrow12, programjames1, Vladimir_Djurica, Adventure10, Mango247

parmenides51 wrote:

The sum of non-negative numbers $x$ , $y$ and $z$ is $3$ . Prove the inequality
${1 \over x ^ 2 + y + z} + {1 \over x + y ^ 2 + z} + {1 \over x + y + z ^ 2} \leq 1.$

Solution. The Cauchy-Schwarz's Inequality gives $(a+b+c)^2\le(a^2+b+c)(1+b+c)$ for any real numbers $a,b,c\ge0$ , and thus
$\begin{align*}&{1 \over x ^ 2 + y + z} + {1 \over x + y ^ 2 + z} + {1 \over x + y + z ^ 2}\\ \leq&\frac{1+y+z}{(x+y+z)^2}+\frac{1+x+z}{(x+y+z)^2}+\frac{1+x+y}{(x+y+z)^2}\\ =&\frac{3+2(x+y+z)}{(x+y+z)^2}=1. \end{align*}$ As required. $\blacksquare$

This post has been edited 1 time. Last edited by ytChen, May 12, 2019, 6:38 PM
Reason: A typo.

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#4 h May 13, 2019, 1:50 PM • 2 Y

Y by Adventure10, Mango247

parmenides51 wrote:

The sum of non-negative numbers $x$ , $y$ and $z$ is $3$ . Prove the inequality
${1 \over x ^ 2 + y + z} + {1 \over x + y ^ 2 + z} + {1 \over x + y + z ^ 2} \leq 1.$

https://artofproblemsolving.com/community/c6h1775336p11678609

For $a, b, c>0, a+b+c=3$ prove that
$\frac{1}{a^2+b+c+1}+\frac{1}{b^2+c+a+1}+\frac{1}{c^2+a+b+1}\le\frac{3}{4}$

This post has been edited 1 time. Last edited by sqing, May 13, 2019, 2:05 PM

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#5 h May 13, 2019, 3:48 PM • 3 Y

Y by programjames1, Adventure10, Mango247

parmenides51 wrote:

The sum of non-negative numbers $x$ , $y$ and $z$ is $3$ . Prove the inequality
${1 \over x ^ 2 + y + z} + {1 \over x + y ^ 2 + z} + {1 \over x + y + z ^ 2} \leq 1.$

Trivial by
$\forall x\leq 3:\ \frac{1}{x^2-x+3}\leq \frac{-x+4}{9}\iff (x-1)^2 (-x+3)\geq 0$

This post has been edited 3 times. Last edited by MrOrange, Sep 17, 2019, 2:21 PM

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#6 h May 13, 2019, 7:04 PM • 1 Y

Y by Adventure10

sqing wrote:

For $a, b, c>0, a+b+c=3$ prove that
$\frac{1}{a^2+b+c+1}+\frac{1}{b^2+c+a+1}+\frac{1}{c^2+a+b+1}\le\frac{3}{4}$

Solution. The Cauchy-Schwarz's Inequality gives $(x+y+z+1)^2\le(x^2+y+z+1)(1+y+z+1)$ for any positive real numbers $x,y,z$ , and thus
$\begin{align*}&\frac{1}{a^2+b+c+1}+\frac{1}{b^2+c+a+1}+\frac{1}{c^2+a+b+1}\\ \leq&\frac{2+b+c}{(a+b+c+1)^2}+\frac{2+c+a}{(a+b+c+1)^2}+\frac{2+a+b}{(a+b+c+1)^2}\\ =&\frac{6+2(a+b+c)}{(a+b+c+1)^2}=\frac{3}{4}. \end{align*}$ As required. $\blacksquare$

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#7 h May 14, 2019, 7:36 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

For $a, b, c>0, a+b+c=3$ prove that
$\frac{1}{a^2+b+c+1}+\frac{1}{b^2+c+a+1}+\frac{1}{c^2+a+b+1}\le\frac{3}{4}$

Also trivial by
$\forall x\leq 4:\ \frac{1}{x^2-x+4}\leq \frac{-x+5}{16}\iff (x-1)^2 (4-x)\geq 0$

This post has been edited 3 times. Last edited by MrOrange, Sep 17, 2019, 2:19 PM

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#8 h May 16, 2019, 12:36 PM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c$ are positive numbers such that $a+b+c=3.$ Prove that $\frac{1}{a^2+k(b+c)} +\frac{1}{b^2+k(c+a)}+\frac{1}{c^2+k(a+b)} \le\frac{3}{2k+1}.$ where $1\leq k\leq 2.$

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#9 h Jun 16, 2022, 12:07 PM

Y by

ytChen wrote:

parmenides51 wrote:

The sum of non-negative numbers $x$ , $y$ and $z$ is $3$ . Prove the inequality
${1 \over x ^ 2 + y + z} + {1 \over x + y ^ 2 + z} + {1 \over x + y + z ^ 2} \leq 1.$

Solution. The Cauchy-Schwarz's Inequality gives $(a+b+c)^2\le(a^2+b+c)(1+b+c)$ for any real numbers $a,b,c\ge0$ , and thus
$\begin{align*}&{1 \over a ^ 2 + b+ c} + {1 \over a + b^ 2 + c} + {1 \over a + b + c ^ 2}\\ \leq&\frac{1+b+c}{(a+b+c)^2}+\frac{1+a+c}{(x+y+z)^2}+\frac{1+x+y}{(x+y+z)^2}\\ =&\frac{3+2(a+b+c)}{(a+b+c)^2}\leq& 1\end{align*}$ Solution. The Cauchy-Schwarz's Inequality gives $(a+b+c)^2\le(a^2+b+c)(1+b+c)$ for any real numbers $a,b,c\ge0$ , and thus
$\begin{align*}&{1 \over a ^ 2 + b+ c} + {1 \over a + b^ 2 + c} + {1 \over a + b + c ^ 2}\\ \leq&\frac{1+b+c}{(a+b+c)^2}+\frac{1+a+c}{(a+b+c)^2}+\frac{1+a+b}{(a+b+c)^2}\\ =&\frac{3+2(a+b+c)}{(a+b+c)^2}\leq&1\end{align*}$

This post has been edited 8 times. Last edited by Nabilos, Jun 16, 2022, 12:20 PM
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