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

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:

$$2abc+3(ab+ac+bc)=27\Leftrightarrow10abc+15(ab+ac+bc)=135$$

so$$135\leq 16(a^2+b^2+c^2)+8abc\Leftrightarrow$$ $$10abc+15(ab+ac+bc)\leq 16(a^2+b^2+c^2)+8abc\Leftrightarrow2abc+15(ab+ac+bc)\leq 16(a^2+b^2+c^2)$$.

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)

• Did you mean "Let's show that $\sqrt{\frac{(a^2+b^2+c^2)^3}{27}}\leq a^2+b^2+c^2$" instead of "Let's show that $2\sqrt{\frac{(a^2+b^2+c^2)^3}{27}}\leq a^2+b^2+c^2$" Commented Jun 23, 2022 at 3:07
• @DevanshBhardwaj the quoted inequality with $\geq135$ at the top of the question is true, but the asker's intermediate workings are not correct Commented Jun 23, 2022 at 3:14
• @Anas, you should fix your question. You clearly did something wrong because your (derived) assertion has a simple counterexample (as exemplified by Ekesh Kumar's post below) but the original proposition is consistent with the counterexample. Which means you obviously did something wrong somewhere, and what you want help with cannot be shown as it is not true. So please either rephrase the question to asking for help with proving the original proposition or retract it entirely. It is unanswerable, especially with your last line saying you don't want help with the general problem. Thank you. Commented Jun 24, 2022 at 16:03
• @Anas, did you simply reverse the sign of the inequality by mistake? Commented Jun 24, 2022 at 16:07
• @Deepak frankly i did not understand what you mean, but the countrexample is for $a^2+b^2+c^2 \leq 27/4$ not for thr original question
– user1069990
Commented Jun 24, 2022 at 19:00

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

• @DevanshBhardwaj It satisfies the condition.
– ling
Commented Jun 23, 2022 at 3:37
• It is a counterexample to the (erroneous) thing that OP wants help with, but not the actual problem statement. Commented Jun 24, 2022 at 16:00

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 $$2\left(\frac{a^2+b^2+c^2}{3}\right)^{\frac{3}{2}}+3(a^2+b^2+c^2)\geq 27$$.)

• nice proof......
– user1069990
Commented Jun 23, 2022 at 11:17

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 $$\overbrace{\ \ \ 2abc\ \ \ }^{\le\,2u^3}+\overbrace{3(ab+bc+ca)}^{9u^2}=27\implies u\ge\frac32\tag2$$

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 $$u\ge\frac32$$

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.

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}

• Would using wolfram alpha for solving the linear equations be considered unacceptable as an answer? Commented Jun 25, 2022 at 12:15
• This looks correct, but it would be nice to explain that $(0,3,3)$ came from solving the two variable Lagrangian.
– robjohn
Commented Jun 25, 2022 at 13:33
• @robjohn sir edited, but I wanted to know what in the answer still attracts the downvote, and what could be improved more. Commented Jun 25, 2022 at 15:37
• I have gotten many mysterious downvotes on what I thought were good answers. The issue might be the advanced level of Lagrange multipliers considering the elementary level of some of the other answers. It could be something unrelated to the content of your answer. People who give these kind of downvotes rarely give a reason. (+1) for the alternative answer, and thanks for editing your answer.
– robjohn
Commented Jun 25, 2022 at 16:53