(Oops, wrong attempt.) Let $x = (ab)^2$, $y = (bc)^2$, $z = (ca)^2$ so that $x + y + z = 3$ and we wish to prove $\prod_{x,y,z} \left(\frac{xz}{y} - \sqrt{\frac{xz}{y}} + 1\right) \geq 1$, i.e. $\prod_{x,y,z} (xz - \sqrt{xyz} + y) \geq xyz$. After expanding we obtain the equivalent $xyz(x^2 + y^2 + z^2) + x^2y^2z^2 + xyz(xy+yz+xz) + xyz(x+y+z) - \sqrt{xyz}(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) - (xyz)^{\frac{3}{2}}(x+y+z) + x^2y^2 + y^2z^2 + x^2z^2 - (xyz)^{\frac{3}{2}} - \sqrt{xyz}(xy + yz + xz) \geq 0$. We have $\frac{xyz}{2}x^2 + \frac{x^2y^2}{2} \geq x^2y\sqrt{xyz}$ and with the analogous ones we obtain $xyz(x^2 + y^2 + z^2) + x^2y^2 + y^2z^2 + z^2x^2 - \sqrt{xyz}(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) \geq 0$. Hence, using also $x+y+z=3$, we reduce to showing $x^2y^2z^2 + xyz(xy+yz+xz) + 3xyz - 4(xyz)^{\frac{3}{2}} - \sqrt{xyz}(xy+yz+xz) \geq 0$. However, $xy + 1 \geq 2\sqrt{xy}$, etc, so $xyz(xy+yz+xz) + 3xyz \geq 2 \sqrt{xyz}(xy+yz+xz)$. It remains to prove that $x^2y^2z^2 - 4(xyz)^{\frac{3}{2}} + \sqrt{xyz}(xy+yz+xz) \geq 0$, but AM-GM yields $xy+yz+xz \geq 3(xyz)^{\frac{2}{3}}$ and hence it suffices to have $t^2 - 4t^{\frac{3}{2}} + 3t^{\frac{7}{6}} \geq 0$ where $t = xyz \leq \left(\frac{x+y+z}{3}\right)^3 = 1$ by AM-GM. The latter is equivalent to $t^{\frac{7}{6}}(t^{\frac{1}{6}} - 1)(t^{\frac{2}{3}} + t^{\frac{1}{2}} + t^{\frac{1}{3}} - 3t^{\frac{1}{6}} - 3) \geq 0$, which is obviously true.

Let $a,b$ be real numbers such that $a^2+a^2b^2+b^2=3.$ Prove that$$(a^2-a+1)(b^2-b+1)\geq 1.$$Let $a,b,c$ be real numbers such that $a^2+ab+b^2\geq 3.$ Prove that$$(a^2-ac+c^2)(b^2-bc+c^2)\geq c^2.$$

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Assassino9931 wrote: Let $x = (ab)^2$, $y = (bc)^2$, $z = (ca)^2$ so that $x + y + z = 3$ and we wish to prove $\prod_{x,y,z} \left(\frac{xz}{y} - \sqrt{\frac{xz}{y}} + 1\right) \geq 1$, i.e. $\prod_{x,y,z} (xz - \sqrt{xyz} + y) \geq xyz$. I think it should be $\prod_{x,y,z} \left(\sqrt{\frac{xz}{y}} - \sqrt[4]{\frac{xz}{y}} + 1\right) \geq 1$.

Thank you!

sqing wrote: Let $a,b$ be real numbers such that $a^2+a^2b^2+b^2=3.$ Prove that$$(a^2-a+1)(b^2-b+1)\geq 1.$$Let $a,b,c$ be real numbers such that $a^2+ab+b^2\geq 3.$ Prove that$$(a^2-ac+c^2)(b^2-bc+c^2)\geq c^2.$$ Nice!

A solution using Schurs and some AM-GM Rewrite the inequality as $$(a^3+1)(b^3+1)(c^3+1)\geq (a+1)(b+1)(c+1)\hspace{20pt}(1)$$Notice that $$(a^2+b^2+c^2)^2\geq 3(a^2b^2+c^2a^2+b^2c^2)=9$$so $a^2+b^2+c^2\geq 3$, therefore, $$a^3+b^3+c^3\geq\frac{(a+b+c)(a^2+b^2+c^2)}{3}\geq a+b+c$$Now expanding $(1)$ it suffices to show $$a^3b^3c^3+a^3b^3+b^3c^3+c^3a^3\geq abc+ab+bc+ca\hspace{20pt}(2)$$Notice that \begin{align*} a^3b^3+b^3c^3+c^3a^3-(ab+bc+ca)&=\frac{3(\displaystyle\sum_{cyc}a^3b^3)-(a^2b^2+b^2c^2+c^2a^2)(ab+bc+ca)}{3}\\ &=\frac{\displaystyle2\sum_{cyc}a^3b^3-\sum_{sym}a^3b^2c}{3}\\ &\geq\frac{a^3b^3+b^3c^3+c^3a^3-3a^2b^2c^2}{3}\end{align*}where the last line follows from Schur's inequality on the variables $ab,bc,ca$. Therefore, by $(2)$ it suffices to show $$a^3b^3+b^3c^3+c^3a^3+3a^3b^3c^3\geq 3a^2b^2c^2+3abc\hspace{20pt}(3)$$Notice that the function $f(x)=x^2+x-x^3$ is strictly increasing in the interval $[0,1]$ since $f'(x)=(1-x)(3x+1)$. By AM-GM $abc\leq 1$, hence $$-3a^3b^3c^3+3a^2b^2c^2+3abc\leq 3f(1)=3$$Meanwhile, by power mean inequality $$\left(\frac{a^3b^3+c^3a^3+b^3c^3}{3}\right)^{\frac{1}{3}}\geq\left(\frac{a^2b^2+c^2a^2+b^2c^2}{3}\right)^{\frac{1}{2}}$$so $a^3b^3+c^3a^3+b^3c^3\geq 3$, this proves $(3)$ and hence finishes the whole problem.

The key observation is $$(a^2-a+1)(b^2-b+1) \geq \frac{a^2+b^2}{2}$$which is equivalent to $\Big( \frac{a+b}{2}-ab \Big)^2 + \Big( \frac{a+b}{2} - 1 \Big)^2 \geq 0$. Using this inequality we have: \[ (a^2 - a + 1)(b^2 - b + 1)(c^2 - c + 1) \geq \sqrt{\frac{(a^2+b^2)(b^2+c^2)(c^2+a^2)}{8}}. \]So, it's enough to show \[ (a^2+b^2)(b^2+c^2)(c^2+a^2) \geq 8. \]Let $a^2=x$, $b^2=y$. and $c^2=z$. Then statement becomes: $xy+yz+zx=3$ and we need to show $$(x+y)(y+z)(z+x) \geq 8$$Since $$(x+y)(y+z)(z+x) \geq \frac{8}{9}(x+y+z)(xy+yz+zx)$$it's just $\sum_{cyc}z(x-y)^2\geq0$. $$=\frac{8}{3}(x+y+z)$$we only need to show $x+y+z \geq 3$. But it's obvious: $x+y+z \geq \sqrt{3(xy+yz+zx)}=3$. Equality holds iff $a=b=c=1$.

Another solution uses $uvw$-method. As in the previous solution, we should prove \[ (x+y)(y+z)(z+x) \geq 8 \]. Let $x+y+z=3u$, $xy+yz+zx=3v^2$, and $xyz=w^3$. After expanding the brackets, we need to show $f(u) \geq 0$, where $f(u)=9u-w^3-8$. Since the function is linear, it will get its minimum value when $u$ gets minimum value. It happens when two numbers are equal, WLOG $y=z$. Then we have: $$x=\frac{3-y^2}{2y}$$and need to show that $$3\Bigg( \frac{3-y^2}{2y}+2y \Bigg)-\frac{y^2(3-y^2)}{2y} \geq 8$$which is equivalent to: $$(y^2-4 \sqrt{y} +3)(y^2+4 \sqrt{y} +3) \geq 0$$ which is true. Equality holds iff $a=b=c=1$.

The official solution uses expansion. Here's a different proof: First we label equations: \begin{align*} a^2b^2 + b^2c^2 + c^2a^2 = 3 \qquad \qquad (1) \\ (a^2 - a + 1)(b^2 - b+1)(c^2 - c+1) \ge 1 \qquad \qquad (\bigstar) \end{align*}From $(1)$, Clearly all $a,b,c$ cannot be $<1$. WLOG assume that $c \ge 1$. Now $(1)$ is equivalent to $$ (a^2 + c^2)(b^2 +c^2) = 3 + c^2 \qquad \qquad (2)$$Pick a $k \in \left( \frac{1}{2} , 2 \right)$ such that $$ c^2 = \frac{ 1 - \frac{k}{2}}{1 - \frac{1}{2k}} \qquad \qquad (3) $$Now for any $x$ we have \begin{align*} x^2 - x + 1 &\ge (x^2 - x + 1) - \frac{(x-k)^2}{2k} \\ &= (x^2 - x + 1) - \frac{x^2 - 2kx + k^2}{2k} \\ &= x^2 \left( 1 - \frac{1}{2k} \right) + \left( 1 - \frac{k}{2} \right) \\ &= \left( 1 - \frac{1}{2k} \right)(x^2 + c^2) \end{align*}Take $x=a,b$. Due to $(2)$, to prove $(\bigstar)$ it suffices to show $$ \left(1 - \frac{1}{2k} \right)^2 (3+c^2)(c^2 - c + 1) \ge 1 \qquad \qquad (\spadesuit) $$ Claim: $$k \le \frac{c+1}{2c}$$ Proof: Assume on the contrary that $k > \frac{c+1}{2c}$. From $(3)$ we obtain \begin{align*} c \le c^2 = \frac{1 - \frac{k}{2}}{1 - \frac{1}{2k}} &< \frac{1 - \frac{c+1}{4c}}{1 - \frac{c}{c+1}} = (c+1) \left( 1- \frac{c+1}{4c} \right) = (c+1) - \frac{(c+1)^2}{4c} \\ &\implies \frac{(c+1)^2}{4c}<1 \end{align*}Which is a contradiction. This proves our Claim. $\square$ Hence to prove $(\spadesuit)$, it suffices to $$ \left(1 - \frac{c}{c+1} \right)^2(3+c^2)(c^2 - c + 1) \ge 1 \qquad \qquad (\heartsuit) $$Now $(\heartsuit)$ is just equivalent to $$ (3+c^2)(c^2 - c +1) \ge (c+1)^2 \qquad \qquad (\diamondsuit)$$But that is true, say because \begin{align*} 3+c^2 &\ge 2(c+1) \iff (c-1)^2 \\ c^2 - c + 1 &\ge \frac{c+1}{2} \iff (c-1)(2c-1) \ge 0 \end{align*}Note here we use $c \ge 1$. This completes the proof. $\blacksquare$

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Let $a,b,c$ be positive real numbers such that $ab+bc+ca= 3$. Prove that $$ (a^2 - a + 1)(b^2 - b + 1)(c^2 - c + 1) \geq 1$$Let $a,b,c$ be positive real numbers such that $a+b+c= 3$. Prove that $$ (a^2 - a + 1)(b^2 - b + 1)(c^2 - c + 1) \geq 1$$

My solution is quite long, using a lot of inequality. First, notice that after multiplying both sides of the inequality by $(a+1)(b+1)(c+1)$ and expanding, it suffices to prove that $$a^3b^3c^3+\sum_{cyc}a^3b^3+\sum_{cyc}a^3\ge abc+ \sum_{cyc}ab+ \sum_{cyc}a$$ Inequality 1. $\frac{1}{3} \sum_{cyc}a^3b^3 \ge 1$ Proof. By Power Mean Inequality, $$\left(\frac{\sum_{cyc}a^3b^3}{3} \right)^{\frac{1}{3}}\ge \left(\frac{\sum_{cyc}a^2b^2}{3} \right)^{\frac{1}{2}}=1$$. This gives the desired inequality. Inequality 2. $\sum_{cyc} a^3 \ge \sum_{cyc} a$ Proof. First we have $$\left( \sum_{cyc} a^2 \right)^2 \ge 3\left(\sum_{cyc}a^2b^2 \right)=9 \Rightarrow \sum_{cyc} (a^2-1) \ge 0$$. By Chebyshev Inequality, $$\sum_{cyc} a(a^2-1)\ge \frac{\left( \sum_{cyc} a\right) \left(\sum_{cyc} (a^2-1) \right)}{3}\ge 0 \Rightarrow \sum_{cyc} a^3 \ge \sum_{cyc} a$$. As desired. Inequality 3. $a^3b^3c^3\ge a^2b^2c^2+abc-1$ Proof. Note that $$(abc-1)^2(abc+1)\ge 0 \Rightarrow a^3b^3c^3\ge a^2b^2c^2+abc-1$$. As desired. Inequality 4. $\frac{2}{3} \sum_{cyc} a^3b^3 + a^2b^2c^2 \ge \sum_{cyc} ab$ Proof. First, notice that $$\left(\sum_{cyc} a^2b^2 \right)\left(\sum_{cyc} ab\right)=3\sum_{cyc} ab \Rightarrow abc\left(\sum_{sym} ab^2\right)=3\sum_{cyc} ab - \sum_{cyc} a^3b^3$$. By Schur Inequality, $$\sum_{cyc} a^3b^3+3a^2b^2c^2\ge abc\left(\sum_{sym} ab^2\right)=3\sum_{cyc} ab - \sum_{cyc} a^3b^3$$$$\Rightarrow 2\sum_{cyc} a^3b^3+3a^2b^2c^2\ge3\sum_{cyc} ab \Rightarrow \frac{2}{3} \sum_{cyc} a^3b^3 + a^2b^2c^2 \ge \sum_{cyc} ab $$. As desired. Now summing all 4 inequalities yields the desired result. We are done.