$ab+ac+bc=2\quad ,\quad\min(10a^2+10b^2+c^2)=?$ 
If $ab+ac+bc=2$, find minimum value of $10a^2+10b^2+c^2$
$1)3\qquad\qquad2)4\qquad\qquad3)8\qquad\qquad4)10$

I used AM-GM inequality for three variables:
$$ab+ac+bc\ge3(abc)^{\frac23}\quad\Rightarrow\quad 2\ge3(abc)^{\frac23}$$
$$10a^2+10b^2+c^2\ge 3(10abc)^{\frac23}$$
Hence $10a^2+b^2+c^2\ge2\times 10^{\frac23}$. But it is not in the options.
 A: Hint: $$10a^2+10b^2+c^2-4(ab+bc+ca)=\frac{2}{5}(5a-b-c)^2+\frac{3}{5}(4b-c)^2$$
A: Suppose $a,b,c>0$. Let
$$ a=\frac1{\sqrt{10}}r\cos\phi\sin\theta,b=\frac1{\sqrt{10}}r\sin\phi\sin\theta,c=r\cos\theta. $$
Then
$$ 10a^2+10b^2+c^2=r^2$$
and
\begin{eqnarray}
&&ab+ac+bc\\
&=&\frac1{20}r^2\sin(2\phi)\sin^2\theta+\frac1{2\sqrt{10}}r^2\cos\phi\sin(2\theta)+\frac1{2\sqrt{10}}r^2\sin\phi\sin(2\theta)\\
&=&\frac1{20}r^2\sin(2\phi)\sin^2\theta+\frac1{2\sqrt{10}}r^2(\cos\phi+\sin\phi)\sin(2\theta) \tag1\\
&\le&\frac1{20}r^2\sin^2\theta+\frac{\sqrt2}{2\sqrt{10}}r^2\sin(2\theta)\tag2\\
&=&r^2\bigg(\frac1{20}\frac{1-\cos(2\theta)}{2}+\frac{1}{\sqrt{20}}\sin(2\theta)\bigg)\\
&=&\frac{1}{40}r^2+r^2\bigg(-\frac{1}{40}\cos(2\theta)+\frac{1}{\sqrt{20}}\sin(2\theta)\bigg) \tag3\\
&\le&\frac{1}{40}r^2+r^2\sqrt{(\frac{1}{40})^2+\frac{1}{20}}\tag4\\
&=&\frac{1}{4}r^2
\end{eqnarray}
from which one has $r^2\ge8$. So
$$ \min(10a^2+10b^2+c^2)=\min r^2=8.$$
Note:

*

*From (1) to (2),
$$ \sin(2\phi)\le 1, \cos\phi+\sin\phi\le \sqrt2 $$
and "=" holds if and only if $\phi=\frac{\pi}{4}$.


*From (3) to (4),
$$ a\cos\theta+b\sin\theta\le\sqrt{a^2+b^2}$$
is used.
