Why you don't post all problem of Turkey JBMO Team Selection Test 2012 in here ?

crazyfehmy wrote: Find the greatest real number $M$ for which \[ a^2+b^2+c^2+3abc \geq M(ab+bc+ca) \] for all non-negative real numbers $a,b,c.$ When $a=b=c$ we must have: $3a^2(a+1-M) \ge 0$ what can be always true just if $M \leq 1$. But when $M=1$, then inequality is equivalent to: $\frac{1}{2} [ (a-b)^2+(a-c)^2+(b-c)^2]+3abc \ge 0$ what is obviously true.

MathUniverse wrote: crazyfehmy wrote: Find the greatest real number $M$ for which \[ a^2+b^2+c^2+3abc \geq M(ab+bc+ca) \] for all non-negative real numbers $a,b,c.$ When $a=b=c$ we must have: $3a^2(a+1-M) \ge 0$ what can be always true just if $M \leq 1$. But when $M=1$, then inequality is equivalent to: $\frac{1}{2} [ (a-b)^2+(a-c)^2+(b-c)^2]+3abc \ge 0$ what is obviously true. Actually I have written the problem wrong, now it is correct.

crazyfehmy wrote: Find the greatest real number $M$ for which \[ a^2+b^2+c^2+3abc \geq M(ab+bc+ca) \] for all non-negative real numbers $a,b,c$ satisfying $a+b+c=4.$ $a=b=2,c=0$ we have $M \le 2$ so we need prove $\sum{a^2}+3abc \ge 2\sum{ab}$ $\iff \sum{a^2}\sum a+12abc \ge 2\sum{ab}\sum a \iff \sum{a^3}+6abc \ge \sum_{sym}{a^2b} $ it is true by use schur inequality

crazyfehmy wrote: Actually I have written the problem wrong, now it is correct. The one you wrote the first was better in my opinion because it is not homogenous. In the second one we use standard method of homogenization to get 3rd degree inequality which can be proved in many ways..

MathUniverse wrote: The one you wrote the first was better in my opinion because it is not homogenous. In the second one we use standard method of homogenization to get 3rd degree inequality which can be proved in many ways.. But it is too easy. This is more suitable for a selection test: Find the greatest real number $M$ such that \[a+b+c+4abc\ge M(ab+bc+ca)\] for all $a,b,c\in [0,1]. Also, for \[a+b+c+3abc\ge M(ab+bc+ca).\]

I think the maximal $M$ is 2 in both cases. we need to prove $a+b+c +3abc \ge 2(ab+ac+bc)$ for $a,b,c \in [0,1]$. if one or two of them equals 0 it is trivial. let assume that $a,b,c \in (0,1]$. let $x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$ where $x,y,z \in [1,\infty)$. then we need to prove $xy+xz+yz+3 \ge 2(x+y+z)$. let $x=u+1,y=v+1,z=w+1$ where $u,v,w \ge 0$. then we have $(u+1)(v+1) + (v+1)(w+1)+(w+1)(u+1)+3 -2(u+v+w+3) = uv+uw+vw \ge 0 $. we have equality when $a=b=c=1$ or $a=0,b=c=1$ and permutations. $M$ must be smaller than 2 in both cases because $a=0,b=c=1$ is still an equality point.

Seems UVW is here killiing all problems. Vasc' one gives if $p,r$ are constant, we have $q$ is maximal when $b=c.$ Original problem eq. with $p^2+3r \ge (M+2)q$ so again $a=b$ or $c=0.$ ( I suppose we can already use this theorem at each competition?)

Vasc wrote: But it is too easy. This is more suitable for a selection test: Find the greatest real number $M$ such that \[a+b+c+4abc\ge M(ab+bc+ca)\] for all $a,b,c\in [0,1]$. Also, for \[a+b+c+3abc\ge M(ab+bc+ca).\] But that inequalities are also easy since function is linear in each variable so we need to check cases when $a,b,c \in \{0,1\}$

The following are not so easy. Find the greatest real number $M$ such that \[ (A)\ \ \ a\sqrt a+b\sqrt b+c\sqrt c+4abc\ge M(ab+bc+ca),\] \[ (B)\ \ \ a\sqrt a+b\sqrt b+c\sqrt c+abc\ge M(ab+bc+ca).\] for all $a,b,c\in [0,1]$ .

Vasc wrote: The following are not so easy. Find the greatest real number $M$ such that \[a\sqrt a+b\sqrt b+c\sqrt c+4abc\ge M(ab+bc+ca)\] for all $a,b,c\in [0,1]$ . I have just an ugly proof for this one. I hope someone will find a nice proof. Inserting $a=b=1, \; c=0$, we get $M\le 2$. We will prove inequality for $M=2$: Just to apply uvw, we make substitution: $(a,b,c)=(x^2,y^2,z^2)$. Using uvw, we need to prove inequality just when two of them are equal (because function is increasing in $w^3$ ). WLOG we can suppose that $a=c$, so we need to prove: $F(a,b)=2a\sqrt{a}+b\sqrt{b}+4a^2b-4ab-2a^2 \ge 0$ $\frac{\partial F}{\partial b}= \frac{3}{2}\sqrt{b}+4a^2-4a$, so $\frac{\partial F}{\partial b}=0$ when $\sqrt{b}=\frac{8}{3}a(1-a)$ Inserting that in $F(a,b)$, it remains to prove: $\left( a^{\frac{3}{2}}(1-a)^2(1+\sqrt{a}) - \frac{27}{128} \right)\sqrt{a}(1-\sqrt{a}) \le 0$ But using Am-Gm, we have: $a\cdot(1-a)\left( \frac{1+\sqrt{a}}{4}\right)^2 \frac{\sqrt{a}}{2}(1-\sqrt{a}) \\ \leq \left(\frac{ a+(1-a) + 2\left( \frac{1+\sqrt{a}}{4}\right)+ \frac{\sqrt{a}}{2}+(1-\sqrt{a})}{6} \right)^6 $ Hence, $a^{\frac{3}{2}}(1-a)^2(1+\sqrt{a}) < 32\cdot \left( \frac{5}{12} \right)^6 < \frac{1}{5}< \frac{27}{128}$ We have proved inequality. Equality holds when $\{a,b,c\}=\{1,1,0\}$. $\blacksquare$

This is also mu solution, MathUniverse! But the calculation in the final part of the proof becomes easy if we use first the inequality \[a(1-a)\le \frac 1{4}.\] In addition, we need to prove also the desired inequality for $a=0$. This is a very easy case.

We can have $M\leq \frac 73$ for $a=b=c=1$.

kunny wrote: We can have $M\leq \frac 73$ for $a=b=c=1$. Yes but the answer is not $\frac{7}{3}.$

These are interesting. If $a,b,c\in [0,1]$, then \[(C) \ \ \ 2(a\sqrt a+b\sqrt b+c\sqrt c)\ge 3(ab+bc+ca-abc);\] \[(D) \ \ \ 3(a\sqrt a+b\sqrt b+c\sqrt c)+\frac{500}{81}abc\ge 5(ab+bc+ca).\]

Vasc wrote: These are interesting. If $a,b,c\in [0,1]$, then \[(C) \ \ \ 2(a\sqrt a+b\sqrt b+c\sqrt c)\ge 3(ab+bc+ca-abc);\] Let $f(x)=2x\sqrt x-3x(b+c-bc)$, then $f'(x)=3\sqrt{x}+3(bc-b-c)$ Note that if $x<(b+c-bc)^2$, $f'(x)<0$, if $x>(b+c-bc)^2$, $f'(x)>0$. Therefore when $x=(b+c-bc)^2$, $f(x)$ gets the minimum value. Easily see that $0\le (b+c-bc)^2 \le 1$. That means if and only if $a=(b+c-bc)^2$, $2(a\sqrt a+b\sqrt b+c\sqrt c)-3(ab+bc+ca-abc)$ gets minimum. As a result, when $2(a\sqrt a+b\sqrt b+c\sqrt c)-3(ab+bc+ca-abc)$ gets minimum, we must have these three equation satisified. $a=(b+c-bc)^2$, $b=(c+a-ca)^2$, $c=(a+b-ab)^2$, which are equivalent to $1-\sqrt{a}=(1-b)(1-c)$, $1-\sqrt{b}=(1-c)(1-a)$, $1-\sqrt{c}=(1-a)(1-b)$ W.L.O.G., let $a=\max(a,b,c)$, then $1-\sqrt{a}\le 1-\sqrt{b}=(1-a)(1-c) \le (1-b)(1-c)=1-\sqrt{a}$. Therefore we must have $a=b$. Similarly we have $a=c$. Sub $a=b=c$ onto the original inequality. The equality case holds when $a=b=c=0$ or $a=b=c=1$

Hey, I tried to solve this one. And I found that the greatest number of M is 7/3 for a=b=c=4/3. What do you think?

konica wrote: Hey, I tried to solve this one. And I found that the greatest number of M is 7/3 for a=b=c=4/3. What do you think? I think you have to think once more.

crazyfehmy wrote: Find the greatest real number $M$ for which \[ a^2+b^2+c^2+3abc \geq M(ab+bc+ca) \] for all non-negative real numbers $a,b,c$ satisfying $a+b+c=4.$

Vasc wrote: These are interesting. If $a,b,c\in [0,1]$, then \[(C) \ \ \ 2(a\sqrt a+b\sqrt b+c\sqrt c)\ge 3(ab+bc+ca-abc)\] If $c=0$ then the inequality is obviously true. Let $a=\frac{1}{(x+1)^2}$, $b=\frac{1}{(y+1)^2}$ and $c=\frac{1}{(z+1)^2}$, where $x$, $y$ and $z$ are non-negative numbers. Hence, we need to prove that $\sum_{cyc}(2x^3y^3+6x^3y^2+6x^3z^2-3x^3yz+3x^3y+3x^3z+18x^2y^2-9x^2yz+$ $+x^3+9x^2y+9x^2z-8xyz+3x^2)\geq0$, which is obvious.

why the answer isnt $\frac{7}{3}$ solution= by schur we have $a^3+b^3+c^3+3abc=6abc+64-12\sum ab\ge \sum ab(a+b)=\sum ab(4-c)=4\sum {ab} - 3abc\Longrightarrow 9abc+64\ge 16\sum ab$ also our given inequality is equivalent to $16+3abc\ge (M+2)\sum ab$ since $(a+b+c)^2\ge 3(ab+bc+ca)\Longrightarrow \frac{16}{3}\ge ab+bc+ca$ adding we get $64+9abc\ge 3(M+3)\sum ab$ comparing we get $M=\frac{7}{3}$

For $c=0$, we must have that $(2+M)a^2 - (8 + 4M)a + 16\geq 0$. Now if $M>2$, this quadratic has solutions $0 < 2\pm \frac{2\sqrt{M^2 - 4}}{M+2} < 4$. So between these two distinct roots we have that this quadratic is negative, contradiction. We will show that $M=2$ indeed work. $$\iff a^2 + b^2 + c^2 + 3abc \geq 2(ab + bc + ca)$$$$\iff (a^2 + b^2 + c^2)(a + b + c) + 12abc \geq 2(ab + bc + ca)(a + b + c)$$$$\iff a^3 + b^3 + c^3 + 12abc + \sum_{sym} a^2b \geq 6abc + 2\sum_{sym} a^2b$$$$\iff a^3 + b^3 + c^3 + 6abc \geq \sum_{sym} a^2b$$This is nothing but well known Schur's Inequality, and so we are done. $\square$

The answer is $M=2$::: $\sum a^2 +3abc \ge \sum ab$ To turn this into Schur==>> Since $a+b+c=4$ $\sum a \sum a ^2 +\sum a^2b +12abc\ge 2\sum a^2b +6abc$ Finally $\sum a^3+6abc\ge \sum a^2b$ done