I tried to solve the below problem, I spend more than 5h just for prove it but without any result , this is the best attempt I did ,just I need to show that $a^2+b^2+c^2\leq \frac{27}{4}$

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> if $a,b,c>0$ and $2abc+3(ab+ac+bc)=27$.  
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> prove that;$16(a^2+b^2+c^2)+8abc\geq 135$
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**My attempt:**

$2abc+3(ab+ac+bc)=27$$\Leftrightarrow$$10abc+15(ab+ac+bc)=135$

so$135\leq 16(a^2+b^2+c^2)+8abc$$\Leftrightarrow$ $10abc+15(ab+ac+bc)\leq 16(a^2+b^2+c^2)+8abc$$\Leftrightarrow$$2abc+15(ab+ac+bc)\leq 16(a^2+b^2+c^2)$.

We know this;$15(ab+bc+ac)\leq 15\sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}=15(a^2+b^2+c^2)$(\*)(schwartz inequality).

and

$2abc\leq 2\sqrt{(\frac{(ab+ac+bc)}{3})^3}\leq 2\sqrt{\frac{(a^2+b^2+c^2)^3}{27}}$.

Let's show that :
$2\sqrt{\frac{(a^2+b^2+c^2)^3}{27}}\leq a^2+b^2+c^2$

$2\sqrt{\frac{(a^2+b^2+c^2)^3}{27}}\leq a^2+b^2+c^2$$\Leftrightarrow$$ \frac{(a^2+b^2+c^2)^3}{27}\leq \frac{(a^2+b^2+c^2)^2}{4}$$\Leftrightarrow$$ \frac{4(a^2+b^2+c^2)^3}{4.27}-\frac{27(a^2+b^2+c^2)^2}{4.27}\leq 0$(\*\*)

If (\*\*) is true then we will have $ 2abc\leq a^2+b^2+c^2$ and that's what we need for complet the proof .

Put$ f(x)=4x^3-27x^2$ ,the only positive solution of this function is$\frac{27}{4}$,and after the variation tableau of this function We can see that $f(x)\leq 0$ for$ x\leq \frac{27}{4}$.

then $ \frac{4(a^2+b^2+c^2)^3}{4.27}-\frac{27(a^2+b^2+c^2)^2}{4.27}\leq 0$,so $ 2abc\leq a^2+b^2+c^2$

$\color{blue}{\textrm{note that $ :a^2+b^2+c^2\leq \frac{27}{4}$}}$

So finally; after (\*) and (\*\*) we can say that :$ 2abc+15(ab+ac+bc)\leq 16(a^2+b^2+c^2)$,and this complet the proof .

the problem is that:i can't show why this $\color{blue}{\textrm{ $ :a^2+b^2+c^2\leq \frac{27}{4}$}}$ is true

So my question is that :can you prove or disprove this:$\color{blue}{\textrm{ $ :a^2+b^2+c^2\leq \frac{27}{4}$}}$ ?.

Note that:( i dont want any answers for the general problem,just i need to develop my attempt if that is possible)