154. Difficult inequality (without derivatives eventually).

by Virgil Nicula, Oct 13, 2010, 9:56 AM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=371558

Prove that $\left\|\begin{array}{c}
x>a>0\\\
y>b>0\\\
\frac ax+\frac by=1\end{array}\right\|\implies$ $\boxed{x^2+y^2\ge a^2+b^2+3ab\cdot\left(\sqrt [3]{\frac ab}+\sqrt [3]{\frac ba}\right)}\ \ge$ $a^2+b^2+6ab\ge \frac 12\cdot\left(\sqrt a+\sqrt b\right)^4$ . Generalization.

Proof 1 (with derivatives). $\frac ax+\frac by=1\iff y=h(x)=\frac {bx}{x-a}$ . Denote $g(x,y)=x^2+y^2$ and $f(x)=$ $g\left(x,\frac {bx}{x-a}\right)=$ $x^2+\left(\frac {bx}{x-a}\right)^2$ , $x>a$ .

Note that $f'(x)=\left[2x+2\cdot\frac {bx}{x-a}\cdot\frac {-ab}{(x-a)^2}\right]\ .s.s.\ \left[(x-a)^3-ab^2\right]\ .s.s.\ \left[x-\left(a+\sqrt [3]{ab^2}\right)\right]\ .s.s.\ \left[x-x_0\right]$ , where $x_0=a+\sqrt [3]{ab^2}$ and

$y_0=h\left(x_0\right)=b+\sqrt [3]{a^2b}$ . Thus, $g(x,y)\ge g\left(x_0,y_0\right)$ $\iff$ $x^2+y^2\ge \left(a+\sqrt [3]{ab^2}\right)^2+\left(b+\sqrt [3]{a^2b}\right)^2=$ $a^2+b^2+3\cdot \left(a\sqrt [3]{ab^2}+b\sqrt [3]{a^2b}\right)$ .

Proof 2 (without derivatives). Observe that $a^2+b^2+3ab\cdot\left(\sqrt [3]{\frac ab}+\sqrt [3]{\frac ba}\right) = \left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^3$ . Thus, $x^2+y^2\ge a^2+b^2+3ab\cdot\left(\sqrt [3]{\frac ab}+\sqrt [3]{\frac ba}\right)$ $\iff$ $x^2+y^2\ge$ $ \left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^3$ $\iff$ $\left(x^2+y^2\right)^{\frac 13}\cdot\left(\frac ax+\frac by\right)^{\frac 23}\ge$ $ a^{\frac{2}{3}}+b^{\frac{2}{3}}$ $\iff$ $\left[\left(x^{\frac 23}\right)^3+\left(y^{\frac 23}\right)^3\right]^{\frac 13}\cdot\left\{\left[\left(\frac ax\right)^{\frac 23}\right]^{\frac 32}+\left[\left(\frac by\right)^{\frac 23}\right]^{\frac 32}\right\}^{\frac 23}\ge$ $ x^{\frac 23}\cdot\left(\frac ax\right)^{\frac 23}+y^{\frac 23}\cdot\left(\frac by\right)^{\frac 23}=$ $a^{\frac 23}+b^{\frac 23}$ , what is a particular case of the


Holder's inequality for $p=3$ and $q=\frac 32$ . Note that $\frac 1p+\frac 1q=1$ .

Generalization. For $\left\|\begin{array}{c}
0\le m<\frac ab\\\\
x>a-mb>0\\\\
y>b>0\\\\
\frac ax+\frac by=1\end{array}\right\|$ and $\frac {a-mb}{x}+\frac by=1$ ascertain the minimum of $g(x,y)\equiv \left[\left(x-my\right)^2+y^2\right]$ . Give

a geometrical interpretation (see
here the Proposed problem from CLIII-message of my mathblog). For $m=0$ obtain the proposed problem.
This post has been edited 44 times. Last edited by Virgil Nicula, Dec 1, 2015, 10:14 AM

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