# Le Khanh Sy's Problem

### Problem

### Solution 1

We first rewrite the inequality as $xa^2+yb^2+zc^2\ge 2m(ab+bc+ca).\,$ Adding $m(a^2+b^2+c^2)\,$ to both sides gives

$(x+m)a^2+(y+m)b^2+(z+m)c^2\ge m(a+b+c)^2.$

Using the Cauchy-Schwarz inequality,

$\displaystyle(x+m)a^2+(y+m)b^2+(z+m)c^2\ge \frac{(a+b+c)^2}{\displaystyle\frac{1}{x+m}+\frac{1}{y+m}+\frac{1}{z+m}}.$

Next, we will show that there is a unique positive real number $m\,$ such that

$\displaystyle\frac{1}{x+m}+\frac{1}{y+m}+\frac{1}{z+m}=\frac{1}{m}$

or,

$\displaystyle 2m^3+(x+y+z)m^2-xyz=0.$

Indeed, consider function $f(m)=2m^3+(x+y+z)m^2-xyz.\,$ Clearly, this function is continuous in the interval $[0,\infty).\,$ Furthermore, $f'(m)=2m(3m+x+y+z)\gt 0,\,$ $f(0)=-xyz\lt 0,\,$ and $\displaystyle\lim_{m\to\infty}=\infty.\,$ Combining these relationships completes the proof.

### Solution 2

The inequality holds for any real $a,b,c\,$ with $ab+bc+ca=1,\,$ and positive $x,y,z.$

Indeed, it's equivalent to

$xa^2-2m(b+c)a+yb^2-2mbc+zc^2\ge 0.$

With, $\Delta\,$ as the discriminant of the quadratic (in $a)\,$ form, we have

$\displaystyle -\frac{\Delta}{4}=(xy-m^2)b^2-2m(x+m)bc+(xz-m^2)c^2.$

On the other hand, the cubic $f(t)=2t^3+(x+y+z)t^2-xyz\,$ is strictly increasing on $(0,\infty).\,$ Now, by the definition, $f(m)=0\,$ and $f(\sqrt{xy})=xy(2\sqrt{xy}+x+y)\gt 0.\,$ So that $\sqrt{xy}\gt m,\,$ i.e., $xy\gt m^2.\,$ Similalrly, $xz\gt m^2.\,$ For the homogeneous quadratic $(xy-m^2)b^2-2m(x+m)bc+(xz-m^2)c^2,\,$ we get

$\displaystyle\begin{align} -\frac{\Delta'}{4}&=(xy-m^2)(xz-m^2)-m^2(x+m)^2\\ &=x[xyz-(x+y+z)m^2]-2m^3x\\ &=2m^3x-2m^3x=0. \end{align}$

The proof is complete.

### Acknowledgment

The problem with the above solution has been posted as a comment to the announcement of Dorin Marghidanu cyclic inequality. Le Khanh Sy posted the problem in Vietnamese and, at the request of professor Marghidanu, Hung Nguyen Viet supplied the translation.

The original problem asked to prove $2a^2+3b^2+4c^2\gt 2\sqrt{2},\,$ provided, $a,b,c\gt 0\,$ and $ab+bc+ca=1.\,$ Observe that, with $x=2,\,$ $y=3,\,$ $z=4,\,$ Le Khanh Sy's method leads to the equation

$2m^3+9m^2-24=0$

whose positive root is approximated at $m\approx 1.4233\gt 1.4142\approx\sqrt{2},\,$ thus strengthening the inequality.

Solution 2 is by Leo Giugiuc.

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