After rewriting the ineq in form of SOS, we have \[3-\sqrt3+\sum\frac{x^2}y-\left(\sum x\right)^2=\sum\frac{(x-y)^2}{y}-\frac12\sum(x-y)^2+\sum x-\sqrt3\\\geq\sum(x-y)^2\left(\frac1y-\frac12\right)\] so $S_y=\frac1x-\frac12,\ldots$ .Let $y$ be middle of $x,z$. note that \[S_y+S_x=\frac 1x+\frac 1z-1\geq z+\frac 1z-1> 0\] and also \[2S_y+S_x=\frac 2x+\frac 1z-\frac32> z+\frac 1z-\frac32> 0\] so the ineq is always true.

goodar2006 wrote: $x,y,z$ are positive real numbers such that $xy+yz+zx=1$. prove that: $3-\sqrt{3}+\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\ge(x+y+z)^2$ (20 points) the exam time was 6 hours. A classical solution using AM-GM inequality : Firstly, we prove the ineq: $ \frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x}\ge\frac{(x+y+z)(x^{2}+y^{2}+z^{2})}{xy+yz+zx}; $ Proof:

So $LHS \ge 3 - \sqrt 3 + (x + y+ z)({x^2} + {y^2} + {z^2})$ And we need to prove: \[\begin{array}{l} 3 - \sqrt 3 + (x + y + z)({x^2} + {y^2} + {z^2}) \ge {(x + y + z)^2} \\ \Leftrightarrow (x + y + z)({x^2} + {y^2} + {z^2}) \ge {x^2} + {y^2} + {z^2} + \sqrt 3 - 1 \\ \Leftrightarrow ({x^2} + {y^2} + {z^2})(x + y + z - 1) \ge \sqrt 3 - 1(1) \\ \end{array}\] Easy to see that ${x^2} + {y^2} + {z^2} \ge xy+yz+zx=1$ and $x +y + z \ge \sqrt{3(xy+yz+zx)}=\sqrt 3$ , so (1) is true we complete the proof, equality holds when $x=y=z= \frac{1}{{\sqrt 3 }}$

goodar2006 wrote: $x,y,z$ are positive real numbers such that $xy+yz+zx=1$. prove that: $3-\sqrt{3}+\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\ge(x+y+z)^2$ (20 points) the exam time was 6 hours. For a similar and stronger problem, visit the following link: Inequalities Marathon . Potla wrote: Problem 370 Given positive reals $a,b,c$ satisfying $ab+bc+ca=1;$ prove that we have \[\frac{a^2}{b}+\frac{b^2}{c}+\frac {c^2}{a}-2(a^2+b^2+c^2)\geq \sqrt 3-2.\]

Potla wrote: Problem 370 Given positive reals $a,b,c$ satisfying $ab+bc+ca=1;$ prove that we have \[\frac{a^2}{b}+\frac{b^2}{c}+\frac {c^2}{a}-2(a^2+b^2+c^2)\geq \sqrt 3-2.\] This ineq can be prove by the ineq i post here: minhhoang wrote: $ \frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x}\ge\frac{(x+y+z)(x^{2}+y^{2}+z^{2})}{xy+yz+zx}; $

we can use Vasc inequality to solve this problem original inequality is \[3-\sqrt3+\sum\frac{x^2}{y}\geq\sum x^2 + 2\] this is equal to \[\sum\frac{x^2}{y}\geq\sum x^2 +\sqrt3-1\] By Vasc inequality \[(\sum\frac{x^2}{\sqrt y})^2\geq\sum 3\frac{x^3}{\sqrt{yz}}\] hence \[\sum x^2\geq\ 3\] enough to show \[\sum\frac{x^3}{\sqrt{yz}}\geq (x^2+y^2+z^2)^2\] But \[\sum\frac{x^3}{\sqrt{yz}}\sum yz\geq \sum x^3\sqrt{x}\sum \sqrt{x} \geq (x^2+y^2+z^2)^2\] so inequality is completed

See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=551109

Using the fact that $xy+yz+zx = 1$, we can rewrite the given inequality as: $$\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} - \sqrt3 \ge x^2 + y^2 + z^2 - 1,$$ or $$x^2 (\frac1{y} - 1) + y^2(\frac1{z}-1) + z^2(\frac1{x}-1) \ge \sqrt3 - 1.$$ Observe that the lists $(\frac1{x} -1, \frac1{y} - 1, \frac1{z} - 1)$ and $(x^2, y^2, z^2)$ are oppositely ordered, and so by Rearrangement we have: $$x^2 (\frac1{y} - 1) + y^2(\frac1{z}-1) + z^2(\frac1{x}-1) \ge y^2(\frac1{y} - 1) + z^2(\frac1{z} - 1) + x^2(\frac1{x}-1). \qquad (1)$$ If all of $x, y, z$ are $\le 1$, then we have from AM-GM that: $$x^2(\frac1{y} - 1) + y^2(\frac1{y} - 1) \ge 2xy(\frac1{y}-1)$$ and similar relations. By $(1)$ and AM-GM we then have that: $$x^2 (\frac1{y} - 1) + y^2(\frac1{z}-1) + z^2(\frac1{x}-1) \ge xy(\frac1{y}-1) + yz(\frac1{z}-1) + zx(\frac1{x}-1) = x+y+z - 1.$$ Since $(x+y+z)^2 \ge 3(xy+yz+zx) = 3$, we've that $x+y+z \ge \sqrt3$ and we're done. Else, we will deal with the case where one of $x, y, z$ is strictly greater than $1$, say $x>1.$ Then it's clear that $y, z \le 1.$ Note that $(1)$ still holds in this case. Hence, in order to finish in the same way as in the previous case, we would just need: $$(x-y)^2(\frac1{y}-1) + (y-z)^2(\frac1{z}-1) + (z-x)^2(\frac1{x}-1) \ge 0,$$ or equivalently $$(x-y)^2(\frac1{y}-1) + (y-z)^2(\frac1{z}-1) \ge (z-x)^2(1-\frac1{x}). \qquad (2)$$ We will show that this is indeed true. If $y \le z$, then $(x-y)^2 > (x-z)^2$, $\frac1{y}-1 \ge 1 - \frac1{x}$ imply that we're done. Else, we have that $x>1>y>z.$ If $x \ge 2$, then $y, z < \frac12$. These imply that $(\frac1{y}-1), (\frac1{z}-1) > 2(1 - \frac1{x}).$ Hence, as $2(x-y)^2 + 2(y-z)^2 \ge [(x-y) + (y-z)]^2$, $(2)$ is proven. Otherwise, suppose that $1 < x < 2.$ In this case, if $(\frac1{y}-1), (\frac1{z}-1) > 2(1 - \frac1{x})$, then we can finish similarly as in the previous paragraph. Otherwise, suppose that $\frac1{y}-1 \le 2(1-\frac1{x}).$ This implies that $y \ge \frac{1}{3 - \frac{2}{x}}$. Since $1 < x < 2$, it's easily checked that this implies $y \ge \frac8{9x}.$ This means that $z < \frac1{x} - y \le \frac{1}{9x}.$ These imply that: $$(y-z)^2 (\frac1{z}-1) >(\frac{7}{9x})^2 (9x-1)^2 > \frac{49}{81x^2} (8x)^2 = \frac{49 \cdot 64}{81} > 4.$$ However, we also have that: $$(z-x)^2(1 - \frac1{x}) < x^2 < 4.$$ Together, these two inequalities prove $(2)$, and so we're done. $\square$