Let $x,y,z$ denote $\cos^2 \alpha, \cos^2 \beta, \cos^2 \gamma$ so we have $x,y,z \geq 0$ and $x+y+z=1$. It suffices to prove that \[2 \leq \sum_{\text{cyc}} (1+x)^2(1-x)^2 \leq (1+x)(1+y)(1+z).\] For the left side, note that $\sum_{\text{cyc}} (1+x)^2(1-x)^2 = \sum_{\text{cyc}} 1-2x^2+x^4$ so it suffices to prove $1-2(x^2+y^2+z^2)+x^4+y^4+z^4 \geq 0.$ However, some algebra yields $x^4+y^4+z^4 + (x+y+z)^4 - 2(x^2+y^2+z^2)(x+y+z)^2$ $= 2(x^2y^2+y^2z^2+z^2x^2)+8xyz(x+y+z) \geq 0.$ For the right side, note that $(x+1)(y+1)(z+1) = xyz+xy+yz+zx+2$ so we need $xyz+xy+yz+zx \geq 1+x^4+y^4+z^4-2(x^2+y^2+z^2).$ Again, we can expand to obtain $xyz(x+y+z)+(xy+yz+zx)(x+y+z)^2$ $+2(x^2+y^2+z^2)(x+y+z)^2- (x^4+y^4+z^4) - (x+y+z)^4$ $= xy(x^2+y^2)+yz(y^2+z^2)+zx(z^2+x^2) - 2xyz(x+y+z) \geq 0$ because $x^3y+xy^3+y^3z+yz^3 \geq 2x^2y^2+2y^2z^2 \geq 4xy^2z$, etc...

It has been discussed before , and I think it's for Austria mo in this form : Let $a,b,c>0 , a+b+c=1$ prove that $2 \leq (1-a^2)^2+(1-b^2)^2+(1-c^2)^2\leq (1+a)(1+b)(1+c)$

Indeed... http://www.kalva.demon.co.uk/aus-pol/apsol/asol009.html

Is there easier solution than this? I think it involves some heavy algebra, especially the $(x+y+z)^4$ term you have to use.

Well, for the right side I used Muirhead. I'll start with the inequality in the form \[2 \leq (1-x^2)^2 + (1-y^2)^2 + (1-z^2)^2 \leq (1+x)(1+y)(1+z)\] with $x+y+z=1$. Factoring we get \[((1-x)(1+x))^2 + ((1-y)(1+y))^2 + ((1-z)(1+z))^2 \leq (1+x)(1+y)(1+z)\] Let $a = 1 - x$, $b = 1 - y$ and $c = 1 - z$. Note that $1 + x = 1 + 1 - y - z = 1 - y + 1 - z = b + c$ and $1 = x + y + z = 1 - a + 1 - b + 1 - c \iff a + b + c = 2$. Thus the inequality is equivalent to \[(a(b+c))^2 + (b(c+a))^2 + (c(a+b))^2 \leq (a+b)(b+c)(c+a)\] Homogenizing and taking symmetric summations we get \[3\sum_{\text{sym}}(a(b+c))^2 \leq \sum_{\text{sym}}(a+b)(b+c)(c+a)\frac{1}{2}(a+b+c)\] \[\iff 6\sum_{\text{sym}}a^2(b^2 + c^2 + 2bc) \leq \sum_{\text{sym}}(6a^2b + 2abc)(a + b + c)\] \[\iff 3\sum_{\text{sym}}(2a^2b^2 + 2a^2bc) \leq \sum_{\text{sym}}(3a^3b + 3a^2b^2 + 6a^2bc)\] \[\iff \sum_{\text{sym}}a^2b^2 \leq \sum_{\text{sym}}a^3b\] Since $[3,1,0]$ majorizes $[2,2,0]$ we are done. The left side is, to me, a bit more interesting (you don't need to think when using Muirhead; now I had to think a lot more). Expanding, we get \[x^4 + y^4 + z^4 + 1\geq 2x^2 + 2y^2 + 2z^2\] Since $x + y + z = 1$, $z = 1 - x - y$, so we have to prove that, for $0 \leq x, y \leq 1$, \[x^4 + y^4 + (1 - x - y)^4 + 1 \geq 2x^2 + 2y^2 + 2(1-x-y)^2\] Expanding again (with binomial expansion formula), we get \[x^4 + y^4 + (x+y)^4 - 4(x+y)^3 + 6(x+y)^2 - 4(x+y) + 2 \geq 2x^2 + 2y^2 + 2 - 4(x+y) + 2(x+y)^2\] \[\iff x^4 + y^4 + (x + y)^4 + 4xy + 2(x+y)^2\geq 4(x+y)^3\] By AM-GM, \[(x+y)^4 + (x+y)^2 \geq 2(x+y)^3\] Then it suffices to prove that \[x^4 + y^4 + x^2 + y^2 + 6xy \geq 2x^3 + 6xy(x+y) + 2y^3\] Again by AM-GM, \[x^4 + x^2 \geq 2x^3\] and \[y^4 + y^2 \geq 2y^3\] so what remains is \[6xy \geq 6xy(x+y)\] or \[x + y \leq 1,\] which is true since $x + y = 1 - z \leq 1$. There must be other shorter ways to do this, but in my opinion this one is natural and relatively straightforward.

correction: it's from the Austrian-Polish math competition

ehsan2004 wrote: It has been discussed before , and I think it's for Austria mo in this form : Let $a,b,c>0 , a+b+c=1$ prove that $$2 \leq (1-a^2)^2+(1-b^2)^2+(1-c^2)^2\leq (1+a)(1+b)(1+c)$$ Austrian-Polish MO 2000 2005 China Northern Math Olympiad:

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