let a=x/2y,b=y/2z,c=z/2x and you get: $$4x^4z^2+4y^4x^2+4z^4y^2+x^4y^2+z^4x^2+y^4z^2 \ge 15x^2y^2z^2$$which is true by am-gm. edit: equality occurs when all the variables are equal i.e. when x=y=z giving a=1/2,b=1/2,c=1/2

apply am-gm to get $a^2+b^2+c^2 \ge 3\sqrt[3]{a^2b^2c^2} = \frac{3}{4}$ and $a^2b^2+b^2c^2+c^2a^2 \ge 3\sqrt[3]{a^4b^4c^4} = \frac{3}{16}$ add them together to get the desired inequality equality occurs when a=b=c=$\frac{1}{2}$

By using AG we get $$\frac{a^2}{4} + \frac{a^2}{4} + \frac{a^2}{4} + \frac{a^2}{4} + \frac{b^2}{4} + \frac{b^2}{4} + \frac{b^2}{4} + \frac{b^2}{4} + \frac{c^2}{4} + \frac{c^2}{4} + \frac{c^2}{4} + \frac{c^2}{4} + a^2b^2+b^2c^2+c^2a^2 \ge 15\sqrt[15]{\frac{a^{12} b^{12} c^{12}}{2^{24}}} =15\sqrt[15]{\frac{1}{2^{60}}}=\frac{15}{16}$$$$ *(abc=\frac{1}{8})$$Equality occurs when a=b=c=1/2

yas jbmo

By AM-GM: \begin{align*} a^2+b^2+c^2&\ge3(abc)^{2/3}=\frac34\\ a^2b^2+b^2c^2+c^2a^2&\ge3(abc)^{4/3}=\frac3{16}\end{align*}and adding these finishes. $\square$

AM - GM nice!

Applying AM-GM for this; a² + b² + c² >= 3/4 = 12/16 a²b² + a²c² + b²c² >= 3/16 And the sum of them are equal to 15/16. We have done)

Also from the 2015 Azerbaijan Junior National. $a ^ 2 + b ^ 2 + c ^ 2 \ge 3(abc)^{2/3} = 3/4, a ^ 2b ^ 2 + b ^ 2c ^ 2 + c ^ 2a ^ 2 \geq 3(abc)^{4/3} = 3/16$, add both to get the answer. Equality of course holds when $a = b = c = \frac{1}{2}$.

parmenides51 wrote: Let $a, b, c$ be positive real numbers such that $abc = \dfrac {1} {8}$. Prove the inequality:$$a ^ 2 + b ^ 2 + c ^ 2 + a ^ 2b ^ 2 + b ^ 2c ^ 2 + c ^ 2a ^ 2 \geq \dfrac {15} {16}$$When the equality holds? Azerbaijan NMO 2015