cjquines0 wrote: Let $x$ and $y$ be positive real numbers such that $x + y = 1$. Show that $$\left( \frac{x+1}{x} \right)^2 + \left( \frac{y+1}{y} \right)^2 \geq 18.$$ $\left( \frac{x+1}{x} \right)^2 + \left( \frac{y+1}{y} \right)^2 =2+\frac{1}{(xy)^2}\geq 18.$

$ \left ( \frac{x+1}{x} \right )^2+ \left( \frac{y+1}{y}\right)^2 \geq \frac{\left( \frac{x+1}{x}+ \frac{y+1}{y} \right)^2}{2}= \frac{ \left(2+ \frac{1}{x}+ \frac{1}{y} \right)^2}{2} \geq \frac{\left(2+ \frac{4}{x+y} \right)^2}{2}= \frac{6^2}{2}=18.$

By AM-GM, we have $1+\frac{1}{x}\geq 5-4x>1$, $\left(1+\frac{1}{x}\right)^2+\left(1+\frac{1}{y}\right)^2\geq (5-4x)^2+(5-4y)^2$ $=16(x^2+y^2)-40(x+y)+50$ $\geq 8(x+y)^2-40(x+y)+50=18$. Equality is attained when $x=y=\frac 12.$

Tangent Line method Let $f(x)=\left(\frac{x+1}{x}\right)^2$,then $f(x)\geq -24x+21$ $\Leftrightarrow$ $(x-\frac{1}{2})^2(24x+4)\geq 0$ This holds when $0 < x < 1$. So $f(x)+f(y)\geq -24+42=18$.

Can anyone correct this plz: By Titu's $(\frac{x+1}{x})^2 + (\frac{y+1}{y})^2 \ge \frac{(x+1+y+1)^2}{x^2+y^2} = \frac{9}{x^2 + y^2}$ but we know $x^2 + y^2 \le \frac{1}{2} $ so the inequality follows.

achen29 wrote: but we know $x^2 + y^2 \le \frac{1}{2} $ so the inequality follows. \[x^2 + y^2 \ge \frac{(x + y)^2}{2} = \frac12\]The inequality is reversed. The Jensen's one is correct.

$(\frac {x+1}{x})^2=(\frac {x+x+y}{x})^2=(2+ \frac {y}{x})^2 =(1+1+\frac {y}{x})^2 \ge 9 (\frac {y}{x})^{\frac {2}{3}} $ Rest is AM GM

Let $x$ and $y$ be positive real numbers such that $x + y = 1$. Show that $$\left( \frac{x+1}{y} \right)^2 + \left( \frac{y+1}{x} \right)^2 \geq 18 $$$$\left( \frac{x^2+1}{y} \right)^2 + \left( \frac{y^2+1}{x} \right)^2\geq \frac{25}{2}$$$$\left( \frac{x+1}{y^2} \right)^2 + \left( \frac{y+1}{x^2} \right)^2 \geq 72$$

Let $f(x) = \left( \frac{x+1}{x} \right)^2$, then $f''(x) = \frac{4x+6}{x^4}>0$ for positive $x$. Now by jensens $$f(x) + f(y) \geq 2 f \left( \frac{x+y}{2} \right) = 2 f(\frac {1}{2}) = 9 \times 2 = 18.$$

sqing wrote: Let $x$ and $y$ be positive real numbers such that $x + y = 1$. Show that $$\left( \frac{x+1}{y} \right)^2 + \left( \frac{y+1}{x} \right)^2 \geq 18 $$$$\left( \frac{x+1}{y^2} \right)^2 + \left( \frac{y+1}{x^2} \right)^2 \geq 72$$ $$xy\leq \frac{1}{4}$$$$\left( \frac{x+1}{y} \right)^2 + \left( \frac{y+1}{x} \right)^2 \geq\frac{2(x+1)(y+1)}{xy}= \frac{2(x+\frac{1}{2}+\frac{1}{2})(y+\frac{1}{2}+\frac{1}{2})}{xy}\geq \frac{18\sqrt[3]{\frac{xy}{16}}}{xy}\geq18 $$

cjquines0 wrote: Let $x$ and $y$ be positive real numbers such that $x + y = 1$. Show that $$\left( \frac{x+1}{x} \right)^2 + \left( \frac{y+1}{y} \right)^2 \geq 18.$$ $$xy\leq \frac{1}{4}$$$$\left( \frac{x+1}{x} \right)^2 + \left( \frac{y+1}{y} \right)^2 \geq\frac{2(x+1)(y+1)}{xy}= \frac{2(x+\frac{1}{2}+\frac{1}{2})(y+\frac{1}{2}+\frac{1}{2})}{xy}\geq \frac{18\sqrt[3]{\frac{xy}{16}}}{xy}\geq18 $$

AlgebraFC wrote:

Splendid

Let $x,y,z$ be positive real numbers such that $x+y+z= 1$. Show that $$ \left( \frac{x+1}{x} \right)^2 + \left( \frac{y+1}{y} \right)^2+ \left( \frac{z+1}{z} \right)^2 \geq 48 $$$$\left( \frac{x+1}{y} \right)^2 + \left( \frac{y+1}{z} \right)^2+ \left( \frac{z+1}{x} \right)^2 \geq 48 $$

By AM-GM, we have $$xy \le \frac{(x+y)^2}{4} = \frac{1}{4}$$so Cauchy-Schwarz gives $$\left( \frac{x+1}{x} \right)^2 + \left( \frac{y+1}{y} \right)^2 = 2 + \frac{2}{x} + \frac{2}{y} + \frac{1}{x^2} + \frac{1}{y^2}$$$$= 2 + \frac{2(x+y)}{xy} + \frac{x^2 + y^2}{(xy)^2} \ge 2 + \frac{2}{xy} + \frac{(x+y)^2}{2 \cdot (xy)^2}$$$$\ge 2 + \frac{2}{\frac{1}{4}} + \frac{1}{2 \cdot \left(\frac{1}{4} \right)^2} = 18$$as desired. Equality holds when $x = \frac{1}{2} = y$. $\blacksquare$

x+y=1 1/2≥√xy 1/16≥x²y² (x+y)²≥16x²y² x²+y²+2xy(x+y)≥16x²y² (x²+y²+2x²y+2y²x)/x²y²≥16 2/x +2/y +1/x² +1/y²≥16 2+ 2/x +2/y +1/x² +1/y²≥18 1/x² +2/x +1 + 1/y² +2/y +1≥ 18 {(x+1)²/x²} + {(y+1)²/y²}≥18