parmenides51 wrote: Show that $\frac{(x + y + z)^2}{3} \ge x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy}$ for all non-negative reals $x, y, z$. Solution. Applying the well-known inequalities $a^2+b^2+c^2\ge ab+bc+ca$ and $(a+b+c)^2\ge3(ab+bc+ca)$, we get \begin{align*}\frac{(x + y + z)^2}{3}=\frac{\left[(\sqrt x)^2 +(\sqrt y)^2 +(\sqrt z)^2\right]^2}{3} \ge\frac{\left(\sqrt{xy} +\sqrt{yz} +\sqrt{zx}\right)^2}{3} \ge x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy}. \end{align*}As stated. $\blacksquare$

parmenides51 wrote: Show that $\frac{(x + y + z)^2}{3} \ge x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy}$ for all non-negative reals $x, y, z$. For $a,b,c >0,$ prove that $$\frac{1}{3}(a+b+c)^2\ge\frac{b+c}{2}\sqrt{bc}+\frac{c+a}{2}\sqrt{ca}+\frac{a+b}{2}\sqrt{ab} \ge\ bc+ ca+ab\ge\sqrt{3abc(a+b+c)}\ge\ a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}.$$

Let's see that (x+y+z)²/3≥xy+yz+zx.Now we should prove xy+yz+zx≥RHS which is true by xy+yz≥2y√xz by AM-GM

KhayalAliyev wrote: Let's see that (x+y+z)²/3≥xy+yz+zx.Now we should prove xy+yz+zx≥RHS which is true by xy+yz≥2y√xz by AM-GM Thanks khayal Aliyev very good solution

İ have something to say this problem was p4 not p3 you have written wrong

parmenides51 wrote: Show that $\frac{(x + y + z)^2}{3} \ge x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy}$ for all non-negative reals $x, y, z$.

Very old idea . By C-S we can say : $(xy+yz+zx)(zx+yx+zy) \geq ( x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy})^2$ Now we know : $$\frac{(x + y + z)^2}{3} \ge xy+yz+zx \geq ( x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy}) $$Now we only need prove $(x+y+z)^2 \geq 3(xy+yz+zx)$ and it is true .

By Rearrangement Inequality, $x.x+y.y+z.z\ge x.y+y.z+z.x$ $x^2+y^2+z^2\ge xy+yz+zx$ Adding $2(xy+yz+zx)$ in both side gives, $(x+y+z)^2\ge3(xy+yz+zx)$ So, $$\frac{(x + y + z)^2}{3} \ge xy+yz+zx$$ By Cauchy-Schwarz inequality, $(xy+yz+zx)(zx+yx+zy) \geq ( x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy})^2$ $xy+yz+zx\geq ( x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy})$ Therefore, $$\frac{(x + y + z)^2}{3} \ge xy+yz+zx\ge ( x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy})$$

$\frac{(\sum_{cyc}x)^2}{3}=\frac{\sum_{cyc}x^2+\sum_{cyc}{xy}}{3}\geq \sum_{cyc}{xy}$. Now as $(1,1,0) \prec (1,1/2,1/2)$, we have by Muirhead, $2(\sum_{cyc}{xy})\geq 2\sum_{cyc}{x\sqrt{yz}}$ which completes our proof.

Let $a=\sqrt{xy}$, $b=\sqrt{yz}$, $c=\sqrt{zx}$. Then the inequality is \[ \left( \frac{ab}{c} \right)^2+\left( \frac{bc}{a}\right)^2+\left( \frac{ca}{b} \right)^2+2(a^2+b^2+c^2) \geq 3(ab+bc+ca)\]Note that \[ \frac{1}{2}\left( \frac{ab}{c} \right)^2+\frac{1}{2}\left( \frac{bc}{a}\right)^2 \geq b^2\]Summing cyclically, we get that \[ \left( \frac{ab}{c} \right)^2+\left( \frac{bc}{a}\right)^2+\left( \frac{ca}{b} \right)^2+2(a^2+b^2+c^2) \geq 3(a^2+b^2+c^2)\]and thus we are done by AM-GM

This inequality is equal to$(x+y+z)^2\geq 3(x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy})$ 1)WLOG $x\geq y\geq z$ $$a_{1}=\sqrt{xy}\hspace{2mm}a_{2}=\sqrt{xz}\hspace{2mm}a_{3}=\sqrt{yz}\hspace{1cm}a_{1}\geq a_{2}\geq a_{3}$$$$b_{1}=\sqrt{xy}\hspace{2mm}b_{2}=\sqrt{xz}\hspace{2mm}b_{3}=\sqrt{yz}\hspace{1cm}b_{1}\geq b_{2}\geq b_{3}$$Using rearrangement inequality, we get:$$xy+xz+yz\geq x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy}$$$$(x+y+z)^2\geq 3xy+3xz+3yz\geq 3(x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy})$$ 2)WLOG $x\geq z\geq y$ $$a_{1}=\sqrt{xz}\hspace{2mm}a_{2}=\sqrt{xy}\hspace{2mm}a_{3}=\sqrt{yz}\hspace{1cm}a_{1}\geq a_{2}\geq a_{3}$$$$b_{1}=\sqrt{xz}\hspace{2mm}b_{2}=\sqrt{xy}\hspace{2mm}b_{3}=\sqrt{yz}\hspace{1cm}b_{1}\geq b_{2}\geq b_{3}$$Using rearrangement inequality, we get:$$xy+xz+yz\geq x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy}$$$$(x+y+z)^2\geq 3xy+3xz+3yz\geq 3(x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy})$$