On the Cartesian plane the curve $(C)$ has equation $x^2=y^3$. A line $d$ varies on the plane such that $d$ always cut $(C)$ at three distinct points with $x$-coordinates $x_1,\, x_2,\, x_3$. a. Prove that the following quantity is a constant: $$\sqrt[3]{\frac{x_1x_2}{x_3^2}}+\sqrt[3]{\frac{x_2x_3}{x_1^2}}+\sqrt[3]{\frac{x_3x_1}{x_2^2}}.$$b. Prove the following inequality: $$\sqrt[3]{\frac{x_1^2}{x_2x_3}}+\sqrt[3]{\frac{x_2^2}{x_3x_1}}+\sqrt[3]{\frac{x_3^2}{x_3x_1}}<-\frac{15}{4}.$$
Problem
Source: Vietnam MO 2018 1st day 4th problem
Tags: analytic geometry, inequalities
quangminhltv99
11.01.2018 10:43
I solved a)
Let $d: y=ax+b$. Then, $x_1,\, x_2,\, x_3$ are the roots of the following equation:
$$
\begin{aligned}
&(ax+b)^3-x^2=0\\
\iff& a^3x^3+(3a^2b-1)x^2+3ab^2x+b^3=0.
\end{aligned}
$$Since $x_1,x_2,x_3$ are distinct, we have $a\ne 0$.
Now, by Vieta, we have:
$$
\begin{aligned}
\sqrt[3]{\frac{x_1x_2}{x_3^2}}+\sqrt[3]{\frac{x_2x_3}{x_1^2}}+\sqrt[3]{\frac{x_3x_1}{x_2^2}}&=\sqrt[3]{\frac{x_1x_2x_3}{x_3^3}}+\sqrt[3]{\frac{x_1x_2x_3}{x_1^3}}+\sqrt[3]{\frac{x_1x_2x_3}{x_2^3}}\\
&=\sqrt[3]{x_1x_2x_3}\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\right)\\
&=\sqrt[3]{x_1x_2x_3}\cdot\frac{x_1x_2+x_2x_3+x_3x_1}{x_1x_2x_3}\\
&=\frac{x_1x_2+x_2x_3+x_3x_1}{\sqrt[3]{(x_1x_2x_3)^2}}\\
&=\frac{\frac{3b^2}{a^2}}{\sqrt[3]{\left(-\frac{b^3}{a^3}\right)^2}}=3.
\end{aligned}
$$
sqing
11.01.2018 11:09
b) $$\sqrt[3]{\frac{x_1^2}{x_2x_3}}+\sqrt[3]{\frac{x_2^2}{x_3x_1}}+\sqrt[3]{\frac{x_3^2}{x_3x_1}}=3-\frac{1}{a^2b}.$$ $$\sqrt[3]{\frac{x_1^2}{x_2x_3}}+\sqrt[3]{\frac{x_2^2}{x_3x_1}}+\sqrt[3]{\frac{x_3^2}{x_3x_1}}<-\frac{15}{4}\iff a^2b<\frac{4}{27}$$
ThE-dArK-lOrD
11.01.2018 20:42
Continue from @above, It's enough to consider only the case when $b>0$. Consider the graph of $f(x)=ax+b-\sqrt[3]{x^2}$, it must intersect $x$-axis at three distinct points. Since $f'(x)=0$ if and only if $x=\frac{8}{27a^3}$, it must be the local minimum (note that $f(0)=b>0$) and so $f(\frac{8}{27a^3})$ must be negative. This gives $b+\frac{8}{27a^2}-\frac{4}{9a^2}<0\implies b<\frac{4}{27a^2}$, done.
quangminh1173
14.01.2018 06:57
b) Let $t=\sqrt[3]{x}$, then $t_1,t_2,t_3$ are distinct roots of the equation: $$at^3-t^2+b=0$$By Vieta, we get: $t_1t_2+t_2t_3+t_3t_1=0 \Rightarrow t_3=-\frac{t_1t_2}{t_1+t_2}$. Using AM-GM, we have: $$\sqrt[3]{\frac{x_1^2}{x_2x_3}}+\sqrt[3]{\frac{x_2^2}{x_3x_1}}+\sqrt[3]{\frac{x_3^2}{x_3x_1}}=\frac{t_1^2}{t_2t_3}+\frac{t_2^2}{t_3t_1}+\frac{t_3^2}{t_3t_1}=-(\frac{t_1^2}{t_2^2}+\frac{t_2^2}{t_1^2}+\frac{t_1}{t_2}+\frac{t_2}{t_1})+\frac{t_1t_2}{(t_1+t_2)^2}<-4+\frac{1}{4}=-\frac{15}{4}.$$
sqing
14.01.2018 07:17
quangminh1173 wrote: b) Let $t=\sqrt[3]{x}$, then $t_1,t_2,t_3$ are distinct roots of the equation: $$at^3-t^2+b=0$$By Vieta, we get: $t_1t_2+t_2t_3+t_3t_1=0 \Rightarrow t_3=-\frac{t_1t_2}{t_1+t_2}$. Using AM-GM, we have: $$\sqrt[3]{\frac{x_1^2}{x_2x_3}}+\sqrt[3]{\frac{x_2^2}{x_3x_1}}+\sqrt[3]{\frac{x_3^2}{x_3x_1}}=\frac{t_1^2}{t_2t_3}+\frac{t_2^2}{t_3t_1}+\frac{t_3^2}{t_3t_1}=-(\frac{t_1^2}{t_2^2}+\frac{t_2^2}{t_1^2}+\frac{t_1}{t_2}+\frac{t_2}{t_1})+\frac{t_1t_2}{(t_1+t_2)^2}<-4+\frac{1}{4}=-\frac{15}{4}.$$ Very very nice. If $x_1,\, x_2,\, x_3$ are the distinct roots of the equation $ (ax+b)^3-x^2=0$ . Then$$\sqrt[3]{\frac{x_1^2}{x_2x_3}}+\sqrt[3]{\frac{x_2^2}{x_3x_1}}+\sqrt[3]{\frac{x_3^2}{x_3x_1}}\leq -\frac{15}{4}.$$