Inequality with a^2+b^2+c^2=1; P4 : B&H TST 2002

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Inequality with a^2+b^2+c^2=1; P4 : B&H TST 2002

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Sayan

#1 h Oct 31, 2014, 10:44 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} \ge \frac35$

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sqing

#2 h Oct 31, 2014, 2:24 PM • 2 Y

Y by ImSh95, Adventure10

$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab}\ge \frac{a^2}{1+b^2+c^2}+\frac{b^2}{1+c^2+a^2}+\frac{c^2}{1+a^2+b^2}$

$=2\left(\frac{1}{2-a^2}+\frac{1}{2-b^2}+\frac{1}{2-c^2}\right)-3\ge \frac{18}{6-(a^2+b^2+c^2)}-3=\frac35$ .

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Sardor

#3 h Nov 1, 2014, 7:12 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

We have by chebishev's inequality and Titu's lemma $3LHS=3 (\frac{a^2}{1+2bc} + \frac{b^2}{1+2ac} + \frac{c^2}{1+2ab}) \ge (a^2+b^2+c^2)(\frac{1}{1+2bc} + \frac{1}{1+2ac} + \frac{1}{1+2ab}) \ge \frac{9}{3+2(ab+bc+ca)} \ge \frac{9}{3+2(a^2+b^2+c^2)} = \frac{9}{5}$ , so we are done.

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sqing

#4 h Sep 5, 2015, 8:24 AM • 2 Y

Y by ImSh95, Adventure10

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a}{1+bc}+\frac{b}{1+ca}+\frac{c}{1+ab} \leq \sqrt{2}.$ Old?

This post has been edited 1 time. Last edited by sqing, Sep 5, 2015, 11:53 AM

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arqady

#5 h Sep 5, 2015, 9:17 AM • 1 Y

Y by Adventure10

sqing wrote:

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a}{1+bc}+\frac{b}{1+ca}+\frac{c}{1+ab} \leq \frac{3\sqrt{3}}{4}.$ Old?

It's wrong! Try $c=0$ and $a=b=\frac{1}{\sqrt2}$ .

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#6 h Sep 5, 2015, 9:37 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

sqing wrote:

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a}{1+bc}+\frac{b}{1+ca}+\frac{c}{1+ab} \leq \frac{3\sqrt{3}}{4}.$ Old?

Here : http://artofproblemsolving.com/community/c6h1138536p5332723

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#7 h Sep 15, 2015, 2:19 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

sqing wrote:

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a}{1+bc}+\frac{b}{1+ca}+\frac{c}{1+ab} \leq \sqrt{2}.$ Old?

For $a,b,c\geq 0$ , we have $1\leq\frac{a}{1+bc}+\frac{b}{1+ca}+\frac{c}{1+ab} \leq\sqrt{2}$
My solution to this:
*Prove the right-hand side inequality:
Note that this following inequality holds for all $x\in [0,\frac{1}{2}]$ :
$\frac{1}{1+x}\leq 1-\frac{2}{3}x$
By AM-GM, $bc\leq\frac{b^2+c^2}{2}\leq\frac{a^2+b^2+c^2}{2}=\frac{1}{2}$
Hence:
$\sum {\frac{a}{1+bc}}\leq\sum a(1-\frac{2}{3}bc)=\sum {a}-2abc$
Now, applying Cauchy-Schwarz inequality yields:
$\sum {a}-2abc=a(1-2bc)+b+c\leq\sqrt{[a^2+(b+c)^2][(1-2bc)^2+1]}$
Note that $a^2+(b+c)^2=1+2bc$ , $(1-2bc)^2+1=2(2b^2c^2-2bc+1)$ .
So:
$\sum {\frac{a}{1+bc}}\leq\sqrt{2[1-2b^2c^2(1-2bc)]}\leq\sqrt{2}$

This post has been edited 2 times. Last edited by huynguyen, Sep 15, 2015, 2:25 PM

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#8 h Nov 29, 2015, 5:44 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Quote:

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} \ge \frac35$

$\sum \frac{a^2}{1+2bc} = \sum \frac{a^2}{ a^2 + \left( b+c \right)^2 } = \sum \frac{ a^4 }{ \left( ab + ac \right)^2 + a^4} \geq \frac{ \left( \sum a^2 \right)^2 } { \sum \left( \left( ab + ac \right)^2 + a^4 \right)} \geq \frac{3}{5}$
Last step follows from the following trivial inequalities ---
$\sum a^2b^2 \geq abc(\sum a)$ , and , $\sum a^4 \geq abc(\sum a)$ .

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#9 h Nov 29, 2015, 6:09 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Sayan wrote:

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} \ge \frac35$

We have
$\sum{\frac{a^2}{a^2+(b+c)^2}}-\frac{3}{5}=\frac{1}{5}\sum{\frac{(a^2+b^2+3c^2+4bc+4ca)(a-b)^2}{[a^2+(b+c)^2][b^2+(c+a)^2]}}\ge{0}$

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#10 h Nov 29, 2015, 8:12 AM • 2 Y

Y by ImSh95, Adventure10

Sayan wrote:

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} \ge \frac35$

$LHS-RHS$
$=\sum_{cyc}{(\frac{a^2}{a^2+(b+c)^2}-\frac{1}{5})}$
$=\sum_{cyc}{\frac{4a^2-(b+c)^2}{5(a^2+(b+c)^2)}}$
$=\sum_{cyc}{\frac{(2a+b+c)((a-b)+(a-c))}{5(a^2+(b+c)^2)}}$
$=\sum_{cyc}{(a-b)S_c}$
Where $S_c=\frac{2a+b+c}{5(a^2+(b+c)^2)} -\frac{a+2b+c}{5(b^2+(a+c)^2)}$
$=(a-b)\frac{a^2+4ac+b^2+4bc+3c^2}{5(a^2+(b+c)^2)(b^2+(a+c)^2)}$
So $LHS-RHS=\sum_{cyc}{(a-b)^2\frac{a^2+4ac+b^2+4bc+3c^2}{5(a^2+(b+c)^2)(b^2+(a+c)^2)}} \geq 0$

This post has been edited 1 time. Last edited by ThE-dArK-lOrD, Nov 29, 2015, 8:13 AM
Reason: Typo

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sqing

#11 h Nov 29, 2015, 9:04 AM • 2 Y

Y by ImSh95, Adventure10

Sayan wrote:

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} \ge \frac35$

Bosnia 2002

http://www.artofproblemsolving.com/community/c6h612114p3640450:
Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} <3.$

This post has been edited 1 time. Last edited by sqing, Nov 29, 2015, 9:18 AM

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#12 h Nov 29, 2015, 9:19 AM • 2 Y

Y by ImSh95, Adventure10

sqing wrote:

Sayan wrote:

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} \ge \frac35$

http://www.artofproblemsolving.com/community/c6h612114p3640450:
Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} <3.$

Inequality is equivalent to $\sum_{cyc}{\frac{a^2}{a^2+(b+c)^2}} <3$
Which is obviously true form $\frac{a^2}{a^2+(b+c)^2} <1$
Summing this with their cyclic permutation.

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41864 posts

sqing

#13 h Nov 29, 2015, 10:21 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Strengthening
Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a^2}{1+2ab}+\frac{b^2}{1+2bc}+\frac{c^2}{1+2ca} \ge \frac35.$

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41864 posts

sqing

#14 h Feb 13, 2020, 12:23 PM • 2 Y

Y by ImSh95, Adventure10

Let $a, b, c > 0$ such that $a^2 + b^2 + c^2 = 3$ . Show that $\frac{1}{a^2 + 2bc} + \frac{1}{b^2 + 2ca} + \frac{1}{c^2 + 2ab} \geq 1$ Let $a, b, c\in [0,1]$ and $a^2 + b^2 + c^2 = 2.$ Show that $1<\frac{a}{1+2bc}+\frac{b}{1+2ca}+\frac{c}{1+2ab}\le 2$ p/6489339085
h

This post has been edited 1 time. Last edited by sqing, Feb 13, 2020, 12:24 PM

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#15 h Jul 29, 2020, 4:56 PM • 1 Y

Y by ImSh95

Using AM-GM $\frac {a^2+b^2+c^2}{3}=\frac {1}{3} \ge (abc)^{2/3} \Rightarrow abc \le \frac {1}{3\sqrt{3}}$ . Also by QM-AM inequality $\frac {a^2+b^2+c^2}{3}=\frac {1}{3} \ge (\frac {a+b+c}{3})^2 \Rightarrow a+b+c \le \sqrt{3} \Rightarrow \frac {1}{3} \ge abc(a+b+c)\Rightarrow \frac {5}{3} \ge 1+2abc(a+b+c)$ . Also $\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} \ge \frac35$ is equivalent to $\sum_{cyc} \frac {a^4}{a^2+2a^2bc} \ge \frac {3}{5}$ . But by Titu's lemma $\sum_{cyc} \frac {a^4}{a^2+2a^2bc} \ge \frac {(a^2+b^2+c^2)^2}{a^2+b^2+c^2+2abc(a+b+c)}\Rightarrow \sum_{cyc} \frac {a^4}{a^2+2a^2bc} \ge \frac {1}{1+2abc(a+b+c)}$ . Combining this with the previous result, we get the desired inequality.

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sqing

#16 h Dec 23, 2020, 12:23 AM • 1 Y

Y by ImSh95

Sayan wrote:

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} \ge \frac35$

$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab}\ge \frac{(a+b+c)^2}{3+2(ab+bc+ca)}=1-\frac{2}{3+2(ab+bc+ca)}\ge 1-\frac{2}{3+2(a^2+b^2+c^2)}=\frac35$ here

This post has been edited 1 time. Last edited by sqing, Dec 23, 2020, 12:24 AM

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#17 h May 5, 2022, 5:17 AM • 1 Y

Y by ImSh95

Note that $\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} \ge \frac{(a^2+b^2+c^2)^2}{a^2 + b^2 + c^2 + 2abc(a+b+c)} = \frac{1}{1+2abc(a+b+c)}$ so we need to prove $5 \ge 3 + 6abc(a+b+c)$ or $1 \ge 3abc(a+b+c)$ . Also Note that $3abc(a+b+c) \le (ab+bc+ca)^2 \le (a^2+b^2+c^2)^2 = 1$ so we're Done.

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MihaiT

#18 h May 5, 2022, 6:07 AM • 1 Y

Y by ImSh95

Mahdi_Mashayekhi wrote:

Note that $\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} \ge \frac{(a^2+b^2+c^2)^2}{a^2 + b^2 + c^2 + 2abc(a+b+c)} = \frac{1}{1+2abc(a+b+c)}$ so we need to prove $5 \ge 3 + 6abc(a+b+c)$ or $1 \ge 3abc(a+b+c)$ . Also Note that $3abc(a+b+c) \le (ab+bc+ca)^2 \le (a^2+b^2+c^2)^2 = 1$ so we're Done.

very nice , but this :

$3abc(a+b+c) \le (ab+bc+ca)^2$ why ,please!

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#19 h Feb 7, 2025, 4:31 PM

Y by

sqing wrote:

Let $a, b, c > 0$ such that $a^2 + b^2 + c^2 = 3$ . Show that $\frac{1}{a^2 + 2bc} + \frac{1}{b^2 + 2ca} + \frac{1}{c^2 + 2ab} \geq 1$

because $ab+bc+ca \leqslant a^2+b^2+c^2$ and by titus we get $\frac{9}{3+2(ab+bc+ca)} \geqslant 1$

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#21 h Feb 7, 2025, 4:37 PM • 1 Y

Y by David_V

MihaiT wrote:

Mahdi_Mashayekhi wrote:

Note that $\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} \ge \frac{(a^2+b^2+c^2)^2}{a^2 + b^2 + c^2 + 2abc(a+b+c)} = \frac{1}{1+2abc(a+b+c)}$ so we need to prove $5 \ge 3 + 6abc(a+b+c)$ or $1 \ge 3abc(a+b+c)$ . Also Note that $3abc(a+b+c) \le (ab+bc+ca)^2 \le (a^2+b^2+c^2)^2 = 1$ so we're Done.

very nice , but this :

$3abc(a+b+c) \le (ab+bc+ca)^2$ why ,please!

Because $(ab+bc+ca)^2 = \sum a^2b^2 + 2\sum a^2bc$ and $a^2b^2 \geqslant a^2bc$

This post has been edited 1 time. Last edited by Juno_34, Feb 7, 2025, 4:37 PM

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#22 h Feb 10, 2025, 2:01 PM • 1 Y

Y by MihaiT

Sayan wrote:

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} \ge \frac35$

by AM-GM we get $\sum \frac{a^2}{1+2bc} \geqslant \sum \frac{a^2}{2-a^2}$ now see that $f(x)=\frac{x}{2-x} -------> f''(x)=\frac{4}{(2-x)^3}$ the function is Convex.
We only need to prove that $f(a^2) + f(b^2) + f(c^2) \geqslant 3 f\left(\frac{a^2+b^2+c^2}{3}\right)$ which is now easy because $a^2+b^2+c^2 = 1.$

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MihaiT

#23 h Feb 11, 2025, 12:40 PM

Y by

Juno_34 wrote:

Sayan wrote:

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$ . Prove that
$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab} \ge \frac35$

by AM-GM we get $\sum \frac{a^2}{1+2bc} \geqslant \sum \frac{a^2}{2-a^2}$ now see that $f(x)=\frac{x}{2-x} -------> f''(x)=\frac{4}{(2-x)^3}$ the function is Convex.
We only need to prove that $f(a^2) + f(b^2) + f(c^2) \geqslant 3 f\left(\frac{a^2+b^2+c^2}{3}\right)$ which is now easy because $a^2+b^2+c^2 = 1.$

beautiful ,! :first:

:first:

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