7
$\begingroup$

Given $a,b,c\ge1;abc\ge8$. Proving that $$\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$$

I have tried by using Jensen's inequality:

We consider the inequality: $\displaystyle\sqrt{x^2-1}\ge\sqrt3+\frac4{\sqrt3}\ln\frac x2\tag{i}$

When $x>\frac85$, I can prove that $(\text i)$ true. But it is clear that $(\text i)$ is not true for all $x<\frac85$, and I can't prove that $\displaystyle\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$ in this case.

$\endgroup$
3
  • $\begingroup$ with jensen,you can have $3\sqrt{\dfrac{a^2+b^2+c^2-3}{3}}$ $\endgroup$
    – chenbai
    Sep 2 '15 at 9:39
  • 1
    $\begingroup$ @chenbai how? $\sqrt{x-1}$ is concave, so the inequality goes in the wrong direction. $\endgroup$ Sep 2 '15 at 9:41
  • $\begingroup$ @barto,ya, you are right .I make a wrong direction. $\endgroup$
    – chenbai
    Sep 2 '15 at 9:45
3
$\begingroup$

$x^2=a^2-1,y^2=b^2-1,z^2=c^2-1 \implies (x^2+1)(y^2+1)(z^2+1)\ge 64 \to x+y+z \ge 3\sqrt{3}$

let $3u=x+y+z,3v^2=xy+yz+xz,w^3=xyz \implies u \ge v \ge w,(x^2+1)(y^2+1)(z^2+1)\ge 64 \iff w^6-6uw^3+9u^2-6v^2+9v^4-63=f(w^3) \ge 0 \implies \Delta\le 0 \implies 36u^2-4(9u^2-6v^2+9v^4-63) \le 0 \iff 3v^4-2v^2-63 \ge 0 \iff (v^2-3)(3v^2+7) \ge 0 \implies v^2 \ge 3 \implies u^2 \ge v^2 \ge 3 \implies x+y+z \ge 3\sqrt{3}$

$\endgroup$
3
$\begingroup$

Set $a=e^x,b=e^y,c=e^z$. Then we have to prove: $$ f(x,y,z)=\sum_{cyc}\sqrt{e^{2x}-1}\geq 3\sqrt{3}\tag{1} $$ with the constraints $x,y,z\geq 0$ and $x+y+z= 3\log 2$. Obviously $x=y=z=\log 2$ is a stationary point (a local minimum) for $f$ over the given domain and $f(\log 2,\log 2,\log 2)=3\sqrt{3}$. The solutions of $$ \frac{e^{2x}}{\sqrt{e^{2x}-1}}=\lambda \tag{2}$$ for $\lambda\geq 2$, are given by $e^x=\frac{1}{\sqrt{2}}\sqrt{\lambda^2\pm \lambda\sqrt{\lambda^2-4}}$, leading to $\sqrt{e^{2x}-1}=\frac{\lambda\pm\sqrt{\lambda^2-4}}{2}$. By studying all the possible chances $(\pm,\pm,\pm)$ (they are not many, up to symmetry) is is not difficult to see that $(x,y,z)=(\log 2)\cdot(1,1,1)$ is a global minimum for the interior of the triangle. After that, we just have to study the boundary of the triangle (one side is enough, always by symmetry) but that is the same problem with just two variables, way easier. By putting things together, we have that the previous point is a global minimum as wanted.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.