Use AM-GM. $a^2+b^2+c^2 \ge 3\sqrt[3]{a^2b^2c^2} = \frac{3}{4}$ and $a^2b^2+b^2c^2+c^2a^2 \ge 3\sqrt[3]{a^4b^4c^4} = \frac{3}{16}$. Summing these two gives the result.

You can also get the following using Hölder's inequality. \begin{align*} \sum _{cyc} a^2 + \sum _{cyc} a^2 b^2 & = (a^2+1)(b^2+1)(c^2+1) - (abc)^2 -1 =(a^2+1)(b^2+1)(c^2+1) -\frac{65}{64} \geq \\ & \left( \sqrt[3]{a^2 b^2 c^2} +1 \right) ^3 -\frac{65}{64} = \left( \frac{5}{4} \right) ^3 - \frac{65}{64} = \frac{15}{16} \end{align*}

IstekOlympiadTeam wrote: Let $a,b$ and $c$ be positive reals such that $abc=\frac{1}{8}$.Then prove that \[a^2+b^2+c^2+a^2b^2+a^2c^2+b^2c^2\ge\frac{15}{16}\] We have: \[P(a,b,c)=a^2+b^2+c^2+a^2b^2+a^2c^2+b^2c^2 \geq a^2+2ab+b^2c^2+2a^2(bc)=P(a,\sqrt{bc},\sqrt{bc})\]So it is enough to prove that: $P(\frac{1}{8t^2},t,t) \geq \frac{15}{16}$ which is true because: \[P(\frac{1}{8t^2},t,t) = \frac{15}{16} +\frac{(2 t - 1)^2 (2 t + 1)^2 (4 t^4 + 10 t^2 + 1)}{(64 t^4)} \geq \frac{15}{16}\]

rkm0959 wrote: Use AM-GM. $a^2+b^2+c^2 \ge 3\sqrt[3]{a^2b^2c^2} = \frac{3}{4}$ and $a^2b^2+b^2c^2+c^2a^2 \ge 3\sqrt[3]{a^4b^4c^4} = \frac{3}{16}$. Summing these two gives the result. Very nice.

I wonder tho.. doesn't this seem a "bit" too straight forward?

Sol:- Let $a^2=x , b^2=y, c^2=z$. We have $xyz=\frac{1}{16}$ and we need to prove that $x+y+z+xy+yz+zx \geq 1-xyz$. Note that $x+y+z+xy+yz+zx \geq 1-xyz \iff x+y+z+xy+yz+zx+xyz+1 \geq 2$ . Using AM-GM we have that $x+y+z+xy+yz+zx+xyz+1 \geq 8(xyz)^{\frac{4}{8}}=2$. So we are done

GeoKing wrote: Sol:- Let $a^2=x , b^2=y, c^2=z$. We have $xyz=\frac{1}{16}$ and we need to prove that $x+y+z+xy+yz+zx \geq 1-xyz$. Note that $x+y+z+xy+yz+zx \geq 1-xyz \iff x+y+z+xy+yz+zx+xyz+1 \geq 2$ . Using AM-GM we have that $x+y+z+xy+yz+zx+xyz+1 \geq 8(xyz)^{\frac{4}{8}}=2$. So we are done -> , because $xyz = \frac{1}{64}$

straight wrote: GeoKing wrote: Sol:- Let $a^2=x , b^2=y, c^2=z$. We have $xyz=\frac{1}{16}$ and we need to prove that $x+y+z+xy+yz+zx \geq 1-xyz$. Note that $x+y+z+xy+yz+zx \geq 1-xyz \iff x+y+z+xy+yz+zx+xyz+1 \geq 2$ . Using AM-GM we have that $x+y+z+xy+yz+zx+xyz+1 \geq 8(xyz)^{\frac{4}{8}}=2$. So we are done -> , because $xyz = \frac{1}{64}$

abc=1/8 a²+b²+c²≥3abc⅔ ≥3/4 a²b²+b²c²+a²c²≥3(a⁴b⁴c⁴)⅓ ≥3/16 So, a²+b²+c²+a²b²+b²c²+a²c²≥3/4+3/16 ≥15/16

Using AM-GM $a^2+b^2+c^2 \ge 3\sqrt[3]{a^2b^2c^2}=\frac{3}{4}$, and we have $a^2b^2+a^2c^2+b^2c^2 \ge 3\sqrt[3]{a^4b^4c^4}=\frac{3}{16}$, adding these up, \[a^2+b^2+c^2+a^2b^2+a^2c^2+b^2c^2\ge\frac{15}{16}\]

The best solution; using AM-GM. First use it for a²+b²+c² we will get a²+b²+c²>=3/4 Then do it for a²b²+b²c²+a²c² and we will get a²b²+b²c²+a²c²>=3/16 Then find the sum of this numbers. 3/4+3/16=12/16+3/16=15/16 it's the answer .

IstekOlympiadTeam wrote: Let $a,b$ and $c$ be positive reals such that $abc=\frac{1}{8}$.Then prove that \[a^2+b^2+c^2+a^2b^2+a^2c^2+b^2c^2\ge\frac{15}{16}\] JBMO Shortlist 2014 A2

IstekOlympiadTeam wrote: Let $a,b$ and $c$ be positive reals such that $abc=\frac{1}{8}$.Then prove that \[a^2+b^2+c^2+a^2b^2+a^2c^2+b^2c^2\ge\frac{15}{16}\] Let $a,b$ and $c$ be positive reals such that $abc=\frac{1}{8}$.Then prove that$$ab+bc+ca+a^2b^2+a^2c^2+b^2c^2\ge\frac{15}{16}$$$$ (a+b+c)^2+a^2b^2+ a^2c^2+b^2c^2\ge\frac{39}{16}$$

That is my first question

$$\sum_{cyc} {a^2} \geq 3\sqrt[3]{a^2b^2c^2} = 3/4$$$$\sum_{cyc} {a^2b^2} \geq 3\sqrt[3]{a^4b^4c^4} = 3/16$$$$\sum_{cyc} {a^2 + a^2b^2}=\sum_{cyc} {a^2} +\sum_{cyc} {a^2b^2} \geq 3/4 + 3/16 = 15/16$$