inequalities (1 Viewer)

untouchablecuz · Oct 23, 2009

$\text {To prove that }(x+\frac{1}{x})^2+(y+\frac{1}{y})^2\geq \frac{25}{2}\\\text {Proven earlier that } xy\leq \frac{1}{4} \text { and given that }x+y=1\\\text {So, }x(1-x)\leq \frac{1}{4}\Rightarrow x+\frac{1}{x}\geq 4-3x \text { after some manipulation}\\\therefore (x+\frac{1}{x})^2\geq (4-3x)^2=16-24x+9x^2\\\text {Similarly, } (y+\frac{1}{y})^2\geq (4-3y)^2=16-24y+9y^2\\\therefore (x+\frac{1}{x})^2+(y+\frac{1}{y})^2\\\geq (16+16)-24(x+y)+9(x^2+y^2)\\=32-24+9((x+y)^2-2xy)\\=8+9(1-2xy)\\\text {But } xy\leq \frac{1}{4} \Rightarrow 1-2xy\geq \frac{1}{2}\\\therefore (x+\frac{1}{x})^2+(y+\frac{1}{y})^2\geq 8+9(1-\frac{1}{2})=\frac{25}{2}$ anyone got a faster way of doing this... only 3 marks

azureus88 · Oct 23, 2009

[maths](x+y)^2=1\\x^2+y^2=1-2xy\geq 1-2(\frac{1}{4})=\frac{1}{2}\\Also,\frac{1}{xy}\geq \frac{1}{4}\Rightarrow \frac{1}{x^2y^2}\geq \frac{1}{16}\\(x+\frac{1}{x})^2+(y+\frac{1}{y})^2\\=x^2+y^2+\frac{x^2+y^2}{x^2y^2}+4\\\geq \frac{1}{2}+\frac{\frac{1}{2}}{\frac{1}{16}}+4\\=\frac{25}{2}[/maths]

untouchablecuz · Oct 23, 2009

beautifullllllll thanks m8

inequalities (1 Viewer)

untouchablecuz

Active Member

azureus88

Member

untouchablecuz

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)