The sides of a triangle have lengths $a,b,c$. Prove that: \begin{align*}(-a+b+c)(a-b+c)\, +\, & (a-b+c)(a+b-c)+(a+b-c)(-a+b+c)\\ &\le\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})\end{align*}
Problem
Source: Romanian TST 2001
Tags: inequalities, inequalities proposed
Akiyama
16.01.2011 18:43
We can prove this problem with Ravi's substitution. Let $a=y+z$, $b=z+x$, $c=x+y$. Then the inequality becomes $4\sum{xy} \le \sqrt{(x+y)(y+z)(z+x)}\sum{\sqrt{x+y}}=\sum{(x+y)\sqrt{(z+x)(z+y)}}$. Since $\sqrt{(z+x)(z+y)} \ge z+\sqrt{xy}$ by Cauchy-Schwartz inequality, $\sum{(x+y)\sqrt{(z+x)(z+y)}} \ge \sum{(x+y)(z+\sqrt{xy})}$. Now, it's enough to prove $\sum{(x+y)\sqrt{xy}} \ge 2\sum{xy}$. This is easy to prove with AM-GM.($x+y \ge 2\sqrt{xy}$) Equality holds if and only of $x=y=z$, that is $a=b=c$.
AK1024
16.01.2011 20:10
Note that $(\sqrt{a}+\sqrt{b}+\sqrt{c})(-\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})=LHS$. Then it is equivalent to proving $(-\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})\le\sqrt{abc}$. This expands to schur's in the third degree, of course there are many other ways of proving it also.
ermal
17.01.2011 21:29
AK1024 wrote: This expands to schur's in the third degree, What does this mean?
MathUniverse
18.01.2011 17:34
ermal wrote: AK1024 wrote: This expands to schur's in the third degree, What does this mean? Schur of third degree is Schur's inequality of degree 3.
We want to prove inequality:
$(a+b-c)(a+c-b)(b+c-a)\le abc$
(You can get @AK 1024's one using $a=\sqrt{a_1}$ and so on..)
First proof: After expanding and using schur's inequality we have:
$abc-(a+b-c)(a+c-b)(b+c-a)$
$=a^3+b^3+c^3-a^2b-a^2c-b^2a-b^2c-c^2a-c^2b + 3abc\ge 0$
Second proof
1.st case: WLOG $c=max\{a,b,c\}$, and $c\ge a+b$. Then $LHS\le 0\le RHS$ and this case is proved.
2.nd case let $a,b,c$ be sides of a triangle. Then we can use substitution $a=x+y$, $b=y+z$ and $c=z+x$ so inequality is equivalent to:
$8xyz\le (x+y)(x+z)(y+z)$
What follows from Am Gm inequality: $u+v \ge 2\sqrt{uv}$
diks94
18.01.2011 19:12
(-a+b+c)(a-b+c) <= c^2 {AM >= GM} L.H.S <= a^2 + b^2 +c^2 also a^2 + b^2 +c^2 >= 3 * (abc)^(2/3) ..1 also R.H.S >= 3* (abc)^(2/3) therfore R.H.S >= L.H.S Hence Proved
ACCCGS8
02.09.2012 10:08
If $a, b, c$ are sides of a triangle, then there is a triangle sides $\sqrt a, \sqrt b, \sqrt c$. So we can make the substitution $a=(x+y)^2, b=(y+z)^2, c=(z+x)^2$ where $x, y, z$ are positive real numbers. Then the problem can easily be finished by expanding and using Muirhead's Inequality.
Nguyenhuyen_AG
02.09.2012 20:44
WakeUp wrote: The sides of a triangle have lengths $a,b,c$. Prove that: \begin{align*}(-a+b+c)(a-b+c)\, +\, & (a-b+c)(a+b-c)+(a+b-c)(-a+b+c)\\ &\le\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})\end{align*} Setting $x=\sqrt{a},\; y=\sqrt{b},\; z=\sqrt{c}.$ We write inequality as \[xyz(x+y+z) \geq (x^2-y^2+z^2)(x^2+y^2-z^2)+(-x^2+y^2+z^2)(x^2+y^2-z^2)+(-x^2+y^2+z^2)(x^2-y^2+z^2),\] or \[xyz(x+y+z)\ge 2(x^2y^2+y^2z^2+z^2x^2)-x^4-y^4-z^4. \quad (1)\] We have \[2(x^2y^2+y^2z^2+z^2x^2)-x^4-y^4-z^4=(x+y+z)(x+y-z)(y+z-x)(z+x-y).\] So, we show that \[xyz(x+y+z)\ge (x+y+z)(x+y-z)(y+z-x)(z+x-y),\] \[xyz\ge (x+y-z)(y+z-x)(z+x-y).\] This is Schur inequality.