prove or disprove that:$a^2+b^2+c^2\leq \frac{27}{4}$

Question

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 $a^2+b^2+c^2\leq \frac{27}{4}$

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

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)

Answer 1

$(a, b, c) = (1, 1, 3)$ is a counterexample.

Answer 2

First by $$a^2+b^2\geq 2ab,\,b^2+c^2\geq 2bc,\,c^2+a^2\geq 2ca$$ we get $$a^2+b^2+c^2\geq ab+bc+ca.$$ Then combining with $2abc+3(ab+bc+ca)=27$ yields $$\begin{aligned} 16(a^2+b^2+c^2)+8abc&=16(a^2+b^2+c^2)+4(27-3(ab+bc+ca))\\ &= 16(a^2+b^2+c^2)-12(ab+bc+ca)+108\\ &\geq 4(a^2+b^2+c^2)+108. \end{aligned}$$ Hence it suffices to prove $a^2+b^2+c^2\geq\frac{27}{4}$. Asuume by contradiction that $a^2+b^2+c^2<\frac{27}{4}$, and note that $a^2+b^2+c^2\geq 3\sqrt[3]{a^2b^2c^2}$, we have $$\begin{aligned}2abc+3(ab+bc+ca)&\leq 2\left(\frac{a^2+b^2+c^2}{3}\right)^{\frac{3}{2}}+3(a^2+b^2+c^2)\\ &<2\left(\frac{9}{4}\right)^{\frac{3}{2}}+3\times\frac{27}{4}\\ &=2\times\frac{27}{8}+3\times\frac{27}{4}=27\end{aligned}$$ contradict with $2abc+3(ab+bc+ca)=27$. Hence $a^2+b^2+c^2\geq\frac{27}{4}$. Then we are done.

(PS1: it is easy to check that " = " holds iff $(a,b,c)=(\frac{3}{2},\frac{3}{2},\frac{3}{2})$.

PS2: We can also get $a^2+b^2+c^2\geq\frac{27}{4}$ by solving the inequality $2\left(\frac{a^2+b^2+c^2}{3}\right)^{\frac{3}{2}}+3(a^2+b^2+c^2)\geq 27$.)

Answer 3

Start with the inequalities $$ \left(\frac{a^2+b^2+c^2}3\right)^{1/2}\overset{\substack{\text{Cauchy}\\\text{Schwarz}\\\downarrow\\[3pt]\\}}{\ge}\overbrace{\left(\frac{ab+bc+ca}3\right)^{1/2}}^u\overset{\substack{\text{AM-GM}\\\downarrow\\[3pt]\\}}{\ge}(abc)^{1/3}\tag1 $$ The given condition implies $$ \overbrace{\ \ \ 2abc\ \ \ }^{\le\,2u^3}+\overbrace{3(ab+bc+ca)}^{9u^2}=27\implies u\ge\frac32\tag2 $$

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Solution to the general problem

Therefore, $$ \begin{align} \hspace{-1cm}16(a^2+b^2+c^2)+8abc &=4\cdot27+16(a^2+b^2+c^2)-12(ab+ac+bc)\tag{3a}\\ &\ge108+4(a^2+b^2+c^2)\tag{3b}\\ &\ge108+12u^2\tag{3c}\\ &\ge135\tag{3d} \end{align} $$ Explanation:
$\text{(3a)}$: $2abc+3(ab+bc+ca)=27$
$\text{(3b)}$: Cauchy-Schwarz says $a^2+b^2+c^2\ge ab+bc+ca$
$\text{(3c)}$: $(1)$ says $a^2+b^2+c^2\ge3u^2$
$\text{(3d)}$: $(2)$ says $u\ge\frac32$

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Solution to the requested problem $$ \begin{align} a^2+b^2+c^2 &\ge3u^2\tag{4a}\\ &\ge\frac{27}4\tag{4b} \end{align} $$ Explanation:
$\text{(4a)}$: apply $(1)$
$\text{(4b)}$: apply $(2)$

That is, the requested inequality is reversed.

Answer 4

Let us call $2abc+3(ab+bc+ac)-27=\phi(a,b,c)$ Let us use the method of Lagrangian Multipliers. \begin{align}\mathcal{L(a,b,c.\lambda)}=16(a^2+b^2+c^2)+8abc+\lambda[2abc+3(ab+bc+ca)-27]\end{align} \begin{align}\nabla \mathcal{L(a,b,c,\lambda)}=0=\left\{ \begin{array}{rcl} 32a+(8+2\lambda)bc+3\lambda(b+c)=0 \\ 32b+(8+2\lambda)ca+3\lambda(c+a)=0 \\ 32c+(8+2\lambda)ab+3\lambda(a+b)=0\\ 2abc+3(ab+bc+ac)-27=0\\ \end{array} \right.\end{align} The solutions to these equations are : \begin{align}(a,b,c)=(-3,-3,\frac{-9}{4}),(\frac{3}{2},\frac{3}{2},\frac{3}{2})(-2.60769,-2.60769,-3.22566)\end{align} (as the inequality and the constraint in symmetric, order of $a,b,c$ does not matter)

Now our additional constraints $a\ge 0,b\ge 0, c\ge 0$, the equation $16(a^2+b^2+c^2)+8abc=r(a,b,c)$ :

Suppose $a=0, b\neq 0, c\neq 0$ from equations 2 and 3, $\frac{32b}{3c}=\frac{32c}{3b}$ unless b and c are zero, and thus $b=\pm c$; however, $b=-c$ leads to imaginary b and c, and so $b=c=3$ \begin{align}r(0,3,3)=288\end{align}

(The constraint equation would not be satisfied for two or more of $a,b,c$ being zero)

So by the theorem of Lagrangian Multipliers, $(\frac{3}{2},\frac{3}{2},\frac{3}{2})$ is a local minimum of $r(a,b,c)$ in the set S where: \begin{align}S=\{(a,b,c)\in\mathbb{R}^3|\phi(a,b,c)=0(a\ge0,b\ge0,c\ge0)\}\end{align} EDIT Just to make it complete, we have: \begin{align}r(\frac{3}{2},\frac{3}{2},\frac{3}{2})=135\end{align}