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inequalities
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Cobedangiu   4
N 5 minutes ago by Cobedangiu
Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$
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Cobedangiu
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Equality occurs in strange points
arqady   10
N Mar 29, 2025 by sqing
Let $a$, $b$ and $c$ are non-negatives such that $a+b+c\neq0$. Prove that:
1) \[\sum_{cyc}\sqrt{a^2+b^2}\leq\frac{11(a^2+b^2+c^2)+7(ab+ac+bc)}{3\sqrt2(a+b+c)}\]
2)\[\sum_{cyc}\sqrt{a^2+7ab+b^2}\leq\frac{2(a^2+b^2+c^2)+7(ab+ac+bc)}{a+b+c}\]
3)\[\sum_{cyc}\sqrt{4a^2+ab+4b^2}\leq\frac{5(a^2+b^2+c^2)+4(ab+ac+bc)}{a+b+c}\]
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arqady
Aug 28, 2011
sqing
Mar 29, 2025
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Equality occurs in strange points
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arqady
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arqady
#1 h  Aug 28, 2011, 11:12 AM • 3 Y
Y by luofangxiang, Adventure10, Mango247
Let $a$, $b$ and $c$ are non-negatives such that $a+b+c\neq0$. Prove that:
1) \[\sum_{cyc}\sqrt{a^2+b^2}\leq\frac{11(a^2+b^2+c^2)+7(ab+ac+bc)}{3\sqrt2(a+b+c)}\]
2)\[\sum_{cyc}\sqrt{a^2+7ab+b^2}\leq\frac{2(a^2+b^2+c^2)+7(ab+ac+bc)}{a+b+c}\]
3)\[\sum_{cyc}\sqrt{4a^2+ab+4b^2}\leq\frac{5(a^2+b^2+c^2)+4(ab+ac+bc)}{a+b+c}\]
This post has been edited 1 time. Last edited by arqady, Aug 29, 2011, 1:03 PM
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Tourish
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Tourish
#2 h  Aug 28, 2011, 11:45 AM • 2 Y
Y by Adventure10, Mango247
I think they are good practice for Cauchy inequality and pqr method :lol:
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arqady
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arqady
#3 h  Aug 28, 2011, 10:28 PM • 2 Y
Y by Adventure10, Mango247
Tourish wrote:
I think they are good practice for Cauchy inequality and pqr method :lol:
What is your proof for the second inequality? I think, it's not so trivially. :wink:
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Tourish
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Tourish
#4 h  Aug 29, 2011, 1:07 AM • 1 Y
Y by Adventure10
For the second one it is very trival if the constant is $9$(it is just your typo?), and in fact I think we have
\[\sum_{cyc}{\sqrt{a^2+7ab+b^2}}\leq\frac{2(a^2+b^2+c^2)+7(ab+bc+ca)}{a+b+c}\]
But I haven't proved it yet :blush:
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arqady
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arqady
#5 h  Aug 29, 2011, 1:04 PM • 2 Y
Y by Adventure10, Mango247
Thank you, Tourish! I have fixed it.
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Tourish
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Tourish
#6 h  Aug 29, 2011, 3:40 PM • 2 Y
Y by Adventure10 and 1 other user
My proof for the second one: if there are two variables of $a,b,c$ are equal to $0$, the original inequality is obvious. Now consider $ab+bc+ca>0$. The inequality is equavilent to
\[\sum{\left(\frac{3}{2}(a+b)-\sqrt{a^2+7ab+b^2}\right)}\geq 3\sum{a}-\frac{2(a^2+b^2+c^2)+7(ab+bc+ca)}{a+b+c}\]
\[\Longleftrightarrow \sum{\frac{5(a-b)^2}{6(a+b)+4\sqrt{a^2+7ab+b^2}}}\geq \frac{\sum{(a-b)^2}}{2(a+b+c)}\]
which can be writen as $S_a(b-c)^2+S_b(c-a)^2+S_c(a-b)^2\geq 0$, where
\[S_c=\frac{5}{6(a+b)+4\sqrt{a^2+7ab+b^2}}-\frac{1}{2(a+b+c)}\]
and $S_a, S_b$ can be expressed similarly. Assume that $a\geq b\geq c$. First, let's prove that $S_b\geq 0$, that is
\[\frac{5}{6(c+a)+4\sqrt{c^2+7ca+a^2}}\geq \frac{1}{2(a+b+c)}\]
\[\Longleftrightarrow 2a+2c+5b\geq 2\sqrt{c^2+7ca+a^2}\]
\[\Longleftrightarrow -20ac+20ab+20bc+25b^2\geq 0\]
which is true since $a\geq b\geq c$. We can prove $S_a\geq 0$ as the same way. Now it suffices to show that $S_b+S_c\geq 0$, which is
\[\frac{5}{6(a+b)+4\sqrt{a^2+7ab+b^2}}+\frac{5}{6(c+a)+4\sqrt{c^2+7ca+a^2}}\geq \frac{1}{a+b+c}\]
By Cauchy-Schwarz inequality we only need to prove
\[\frac{20}{12a+6b+6c+4\sqrt{a^2+7ab+b^2}+4\sqrt{c^2+7ca+a^2}}\geq \frac{1}{a+b+c}\]
\[\Longleftrightarrow 4a+7b+7c\geq 2(\sqrt{a^2+7ab+b^2}+\sqrt{c^2+7ca+a^2})\]
By Cauchy-Schwarz again, we need to prove that
\[(4a+7b+7c)^2\geq 8(2a^2+b^2+c^2+7ab+7ac)\]
which is obvious true after expanding. Then by SOS theory, our inequality is true. Equal holds only when $a=b=c$ or $a=b=0$ and its permutations .
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can_hang2007
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can_hang2007
#7 h  Sep 1, 2011, 5:26 AM • 5 Y
Y by Adventure10, Mango247, MS_asdfgzxcvb, and 2 other users
arqady wrote:
Let $a$, $b$ and $c$ are non-negatives such that $a+b+c\neq0$. Prove that:
2)\[\sum_{cyc}\sqrt{a^2+7ab+b^2}\leq\frac{2(a^2+b^2+c^2)+7(ab+ac+bc)}{a+b+c}\]
I will post my solution to this one.

Without loss of generality, assume that $a \ge b \ge c.$ If $b=c=0,$ then the inequality is trivial, so we only need to consider $b>0.$ Now, using the AM-GM inequality, we get
$2\sqrt{a^2+7ab+b^2} \le a+2b+\frac{a^2+7ab+b^2}{a+2b}=2a+2b+\frac{5ab+b^2}{a+2b},$
$2\sqrt{a^2+7ac+c^2} \le a+2c+\frac{a^2+7ac+c^2}{a+2c} =2a+2c+\frac{5ac+c^2}{a+2c}$
and
$2\sqrt{b^2+7bc+c^2} \le b+2c+\frac{b^2+7bc+c^2}{b+2c}=2b+2c+\frac{5bc+c^2}{b+2c}.$
Therefore, it suffices to prove that
$\frac{5ab+b^2}{a+2b}+\frac{5ac+c^2}{a+2c}+\frac{5bc+c^2}{b+2c} \le \frac{4(a^2+b^2+c^2)+14(ab+bc+ca)}{a+b+c}-4(a+b+c),$
or
$\frac{5ab+b^2}{a+2b}+\frac{5ac+c^2}{a+2c}+\frac{5bc+c^2}{b+2c} \le \frac{6(ab+bc+ca)}{a+b+c}.$
Multiplying each side by $2$ with noting that $\frac{2(5ab+b^2)}{a+2b}=b+\frac{9ab}{a+2b},$ we can rewrite the above inequality as
$b+2c+\frac{9ab}{a+2b}+\frac{9ac}{a+2c}+\frac{9bc}{b+2c} \le \frac{12(ab+bc+ca)}{a+b+c},$
or
$(b+2c)(a+b+c)+\frac{9ab(a+b+c)}{a+2b}+\frac{9ac(a+b+c)}{a+2c}+\frac{9bc(a+b+c)}{b+2c}\le $
$\le 12(ab+bc+ca).$
This rewrites as
$9a\left[\frac{b(a+b+c)}{a+2b}+\frac{c(a+b+c)}{a+2c}- b-c\right] +(a+b+c)\left(\frac{9bc}{b+2c} -2b-c\right) \le $
$\le 12(ab+bc+ca)- 9a(b+c)-(2b+c)(a+b+c)-(b+2c)(a+b+c).$
After some simple computations, we get
$-\frac{9a^2(b-c)^2}{(a+2b)(a+2c)} -\frac{2(a+b+c)(b-c)^2}{b+2c} \le -3(b-c)^2, $
or
$(b-c)^2\left[\frac{9a^2}{(a+2b)(a+2c)}+\frac{2(a+b+c)}{b+2c} -3\right] \ge 0,$
which is true because $\frac{9a^2}{(a+2b)(a+2c)} \ge 1$ and $\frac{2(a+b+c)}{b+2c} \ge 2.$
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luofangxiang
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luofangxiang
#8 h  Dec 4, 2016, 1:18 PM • 2 Y
Y by Adventure10, Mango247
arqady wrote:
Let $a$, $b$ and $c$ are non-negatives such that $a+b+c\neq0$. Prove that:
1) \[\sum_{cyc}\sqrt{a^2+b^2}\leq\frac{11(a^2+b^2+c^2)+7(ab+ac+bc)}{3\sqrt2(a+b+c)}\]2)\[\sum_{cyc}\sqrt{a^2+7ab+b^2}\leq\frac{2(a^2+b^2+c^2)+7(ab+ac+bc)}{a+b+c}\]3)\[\sum_{cyc}\sqrt{4a^2+ab+4b^2}\leq\frac{5(a^2+b^2+c^2)+4(ab+ac+bc)}{a+b+c}\]

//cdn.artofproblemsolving.com/images/8/a/b/8ab4482f5069c0a532962fed49e9a543a5313ea3.jpg
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luofangxiang
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luofangxiang
#9 h  Dec 4, 2016, 1:28 PM • 2 Y
Y by Adventure10, Mango247
（1）//cdn.artofproblemsolving.com/images/8/9/e/89ec700132c4078994bb73bf252d9f2ecdc7f98d.jpg
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manutd03
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manutd03
#10 h  Aug 1, 2018, 8:28 AM • 1 Y
Y by Adventure10
Can we solve the first one by S.O.S theorem, i think i got some trouble with Sa,Sb,Sc
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sqing
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sqing
#11 h  Mar 29, 2025, 1:55 PM
Y by
can_hang2007 wrote:
*
$\frac{5ab+b^2}{a+2b}+\frac{5ac+c^2}{a+2c}+\frac{5bc+c^2}{b+2c} \le \frac{6(ab+bc+ca)}{a+b+c}.$
**
Let $ a,b,c\ge 0 $ and $ ab+bc+ca>0 .$ Prove that
$$\frac{a^2+5bc}{b+c}+\frac{b^2+5ca}{c+a} +\frac{c^2+5ab}{a+b}\le \frac{9(ab+bc+ca)}{a+b+c} $$
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