How to prove $|a^{2} + b^{2} + c^{2} - 2\left( ab+bc+ac\right)| \le \frac{1}{27}$

Question

Let $a,b,c$ are $3$ edge of a triangle and $\sqrt{a} + \sqrt{b} + \sqrt{c} = 1.$

How to prove $|a^{2} + b^{2} + c^{2} - 2\left( ab+bc+ac\right)| \le \frac{1}{27}$? Can this be proved with simple way?

Answer 1

Yes , this can be proved using Lagrange's multiplier.. Let $$f(a,b,c)=a^2+b^2+c^2-2(ab+bc+ca)+k(\sqrt{a}+\sqrt{b}+\sqrt{c}-1)$$ Solve the following four equations: $$\frac{df}{da}=0$$ $$\frac{df}{db}=0$$ $$\frac{df}{dc}=0$$ $$ \sqrt{a}+\sqrt{b}+\sqrt{c}=1$$

You will get a set of values for $(a,b,c)$ Put these values of (a,b,c) in $f$

You get the minimum and maximum values

Answer 2

Let $a=\frac{y+z}{18}$, $b=\frac{x+z}{18}$ and $c=\frac{x+y}{18}$.

Hence, $x$, $y$ and $z$ are positives, $\sum\limits_{cyc}\sqrt{x+y}=3\sqrt2$ and we need to prove that $xy+xz+yz\leq3$, which we can prove by the Contradiction method.

Let $x=ku$, $y=kv$ and $z=kw$, where $k>0$ and $uv+uw+vw=3$.

Hence, $k>1$ and $3\sqrt2=\sum\limits_{cyc}\sqrt{x+y}=\sqrt{k}\sum\limits_{cyc}\sqrt{u+v}>\sum\limits_{cyc}\sqrt{u+v}$,

which is contradiction because we'll prove now that $\sum\limits_{cyc}\sqrt{u+v}\geq3\sqrt2$.

Indeed, we need to prove that

$2(u+v+w)+2\sum\limits_{cyc}\sqrt{(u+v)(u+w)}\geq6\sqrt{3(uv+uw+vw)}$ or

$6(u+v+w)-6\sqrt{3(uv+uw+vw)}\geq\sum\limits_{cyc}\left(\sqrt{u+w}-\sqrt{v+w}\right)^2$ or

$\sum\limits_{cyc}\frac{3(u-v)^2}{u+v+w+\sqrt{3(uv+uw+vw)}}\geq\sum\limits_{cyc}\frac{(u-v)^2}{\left(\sqrt{u+w}+\sqrt{v+w}\right)^2}$, which is obvious because

$3\left(u+v+2w+2\sqrt{w^2+uv+uw+vw}\right)\geq u+v+w+\sqrt{3(uv+uw+vw)}$.

Done

Answer 3

$L=\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)\left(\sqrt{a}-\sqrt{b}+\sqrt{c}\right)\left(-\sqrt{a}+\sqrt{b}+\sqrt{c}\right)=\\ \left(1-2\sqrt{a}\right)\left(1-2\sqrt{b}\right)\left(1-2\sqrt{c}\right)\le\left(\frac{3-2(\sqrt{a}+\sqrt{b}+\sqrt{c})}{3}\right)^3=R$