Posted here http://www.mathlinks.ro/Forum/viewtopic.php?t=71330

After applying Rolle's theorem to $(x-a)(x-b)(x-c)(x-d)$ we get an inequality in $u,v,w$ s.t. $u+v+w=0$. (technique similar with the one used for proving Schur for several variables) I solved that through $w=-(u+v)$; then I considered $LHS-RHS$ as $f(u)$; $f' \left( - \frac{v}{2} \right)=0$; the rest is easy. Can anyone find a nice solution for the inequality with $u+v+w=0$: \[ (uv+vw+wu)^2 + 3 \geq 6uvw \ ? \]

Perfect_Radio, can you post your solution in detail?

We have $P(x) = x^4+(ab+ac+ad+bc+bd+cd)x^2-x(abc+abd+acd+bcd)+abcd$, so $P'(x) = 4x^3 + 2(ab+ac+ad+bc+bd+cd)x-(abc+abd+acd+bcd)$. By Rolle's Theorem, there are $u,v,w \in \mathbb{R}$ s.t. $P'(x) = 4(x-u)(x-v)(x-w)$. (in case we have $a=b$ for example, then $a$ is also a root $P'(x)$; else we apply the Rolle argument) Now equate the two formulas and substitute in the inequality. As I said in my previous post, my solution for the $u+v+w=0$ inequality is pretty ugly (you can do it yourself). Try to solve this inequality. Maybe you'll find a nice solution.

perfect_radio wrote: Can anyone find a nice solution for the inequality with $u+v+w=0$: \[ (uv+vw+wu)^2 + 3 \geq 6uvw \ ? \] You can easy see that we only need to prove this ineq. $x^2y^2+(x^2+y^2)(x+y)^2+3\geq 6xy(x+y)$wher $x,y\geq 0$. Let's use Caush's ineq. for 12 numbers $x^2y^2+8*\frac{(x^2+y^2)(x+y)^2}{8}+1+1+1\geq \sqrt[12]{x^2y^2*(\frac{(x^2+y^2)(x+y)^2}{8})^8}\geq 6xy(x+y)$ Mybe not wery nice but not wery ugly too .

perfect_radio wrote: Can anyone find a nice solution for the inequality with $u+v+w=0$: \[ (uv+vw+wu)^2 + 3 \geq 6uvw \ ? \] Suppose $u \geq 0 \geq v \geq w$ Put $m=-v$ ; $n=-w$ So $u =m+n$ (1) Replace (1) into the ineq , we need to prove : $(m^n+n^2+mn)^2 +3 \geq 6mn(m+n)$ , where $n \geq m \geq 0$ Maybe this ineq can be solved by AM-GM , Cauchy ,... but I used calculus to solve this ineq . It's not nice but easy .

nttu wrote: perfect_radio wrote: Can anyone find a nice solution for the inequality with $u+v+w=0$: \[ (uv+vw+wu)^2 + 3 \geq 6uvw \ ? \] Suppose $u \geq 0 \geq v \geq w$ Put $m=-v$ ; $n=-w$ So $u =m+n$ (1) Replace (1) into the ineq , we need to prove : $(m^n+n^2+mn)^2 +3 \geq 6mn(m+n)$ , where $n \geq m \geq 0$ Maybe this ineq can be solved by AM-GM , Cauchy ,... but I used calculus to solve this ineq . It's not nice but easy . I think you have a typo ther must be $(m^2+n^2+mn)^2 +3 \geq 6mn(m+n)$ ,and for this ineq. see my laste post.(change m,n->x,y )

\displaystyle d=-a-b-c \displaystyle abc+abd+acd+cbd=abc-(a+b+c)(ab+bc+ac)=-(-abc+\sum_{sym}^{}a^2b+3abc)=-(\sum_{sym}^{}a^2b+2abc)=-(a+b)(a+c)(b+c) \displaystyle ab+ac+bc+ad+cd+bd=ab+bc+ac-(a+b+c)^2=-\frac{1}{2}((a+b)^2+(a+c)^2+(b+c)^2) Let \displaystyle x=a+b y=a+c z=b+c We need \displaystyle \frac{1}{4}(x^2+y^2+z^2)^2+12\ge -6xyz \displaystyle x^2+y^2+z^2\ge 3|xyz|^{\frac{2}{3}} So, \displaystyle \frac{1}{4}(x^2+y^2+z^2)^2+12\ge \frac{9}{4}|xyz|^{\frac{4}{3}}+12=\frac{3}{4}|xyz|^{\frac{3}{4}}+\frac{3}{4}|xyz|^{\frac{3}{4}}+\frac{3}{4}|xyz|^{\frac{3}{4}}+12\ge 4\times(\frac{12\times3^3}{4^3})^{\frac{1}{4}}|xyz|=6|xyz|\ge-6xyz

$ d=-a-b-c$ $ abc+abd+acd+cbd=abc-(a+b+c)(ab+bc+ac)=-(-abc+\sum_{sym}^{}a^2b+3abc)=-(\sum_{sym}^{}a^2b+2abc)=-(a+b)(a+c)(b+c)$ $ ab+ac+bc+ad+cd+bd=ab+bc+ac-(a+b+c)^2=-\frac{1}{2}((a+b)^2+(a+c)^2+(b+c)^2)$ Let $ x=a+b$ $ y=a+c$ $ z=b+c$ We need $ \frac{1}{4}(x^2+y^2+z^2)^2+12\ge -6xyz$ $ x^2+y^2+z^2\ge 3|xyz|^{\frac{2}{3}}$ So, $ \frac{1}{4}(x^2+y^2+z^2)^2+12\ge$ $ \frac{9}{4}|xyz|^{\frac{4}{3}}+12=\frac{3}{4}|xyz|^{\frac{4}{3}}+\frac{3}{4}|xyz|^{\frac{4}{3}}+\frac{3}{4}|xyz|^{\frac{4}{3}}+12\ge$ $ 4\times(\frac{12\times3^3}{4^3})^{\frac{1}{4}}|xyz|=6|xyz|\ge-6xyz$

As a three-year late side note, we can prove the inequality in $ u,v,w$ by using the fact $ x^3 + ax + b$ has three real roots iff \[ 4 a^3 + 27 b^2 \leq 0 . \]

Valentin Vornicu wrote: Let $ a,b,c,d$ be real numbers with sum 0. Prove the inequality: \[ (ab + ac + ad + bc + bd + cd)^2 + 12\geq 6(abc + abd + acd + bcd).\] My soloution is here http://vimf.freeforums.org/m-t-s-bdt-olympiad-t162-20.html

litbon wrote: Valentin Vornicu wrote: Let $ a,b,c,d$ be real numbers with sum 0. Prove the inequality: \[ (ab + ac + ad + bc + bd + cd)^2 + 12\geq 6(abc + abd + acd + bcd).\] My soloution is here http://vimf.freeforums.org/m-t-s-bdt-olympiad-t162-20.html Next time just post the solution here. it seems your forum is out of service now..

Can anyone find a nice solution for the inequality with $u+v+w=0$: \[ (uv+vw+wu)^2 + 3 \geq 6uvw \ ? \][/quote] Substituting $w=-(u+v)$, we have the inequality $(u^2+uv+v^2)^2+3\geq6uv(u+v)$ After writing $u=z+t, v=z-t$, the inequality becomes $(3z^2+t^2)^2+ 3 \geq 12z(z^2-t^2)$ As $t^2 \geq 0$ , it is sufficient to prove that $(3z^2)^2+3 \geq 12z^3$, which is obvious from AM-GM

Does anyone find the a,b,c,d when LHS and RHS are equal?

hi can any one help me with this problem i ve seen seen the aops solution but didnt understand we have 0<=x,y,z<=1 prove that (x/(y+z+1))+(y/(x+z+1))+(z/(x+y+1))+(1-x)(1-y)(1-z)<=1

Valentin Vornicu wrote: Let $ a,b,c,d$ be real numbers with sum 0. Prove the inequality: \[ (ab + ac + ad + bc + bd + cd)^2 + 12\geq 6(abc + abd + acd + bcd). \] Let $4u=a+b+c+d$, $6v^2=ab+cd+ac+bd+ad+bc$ ($v^2$ can be negative), $4w^3=abc+bcd+cda+dab$ and $P(x)=(x-a)(x-b)(x-c)(x-d)=x^4-4ux^3+6v^2x^2-4w^3x+abcd$. Thus by Rolle's theorem $P'(x)=4(x^3-3ux^2+3v^2x-w^3)$ has a three real roots, which gives $$\Delta=6912(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=-6912(4v^6+w^6)\geq 0\text{.}$$So we need to prove $$3v^4+1\geq 2w^3$$,after squaring it's equivalent to $$9v^8+6v^4+1\geq 4w^6$$or $$9v^8+6v^4-4w^6+1\geq 9v^8+16v^6+6v^4+1=(3v^2+1)^2(9v^4-2v^2+1)\geq 0$$, which is true.