Question

$\red{ \rm Prove \: that \: ∀ \: x,y \in \R, the } \\ \rm \color{green}{following \: inequality \: hold : } \\ \color{blue} \rm {x}^{2} + {y}^{2} \geq 2xy$​

Answer 1

$\large\underline{\sf{Solution-}}$

As it is given that

$\rm \: x,y \: \in \: R$

$\rm\implies \: {x}^{2}, {y}^{2} \: \in \: R$

Now, Arithmetic mean (AM) between x² and y² is given by

$\rm \: AM = \dfrac{ {x}^{2} + {y}^{2} }{2} - - - (1)$

Now, Geometric mean (GM) between x² and y² is given by

$\rm \: GM = \sqrt{ {x}^{2} \times {y}^{2} } = xy - - - (2)$

As we know that,

$\rm \: AM \geqslant GM$

So, on substituting the values from equation (1) and (2), we get

$\rm \: \dfrac{ {x}^{2} + {y}^{2} }{2} \geqslant xy$

$\color{green}\rm\implies \: {x}^{2} + {y}^{2} \geqslant 2xy$

$\rule{190pt}{2pt}$

Additional Information :-

Relationship between Arithmetic mean, Geometric mean and Harmonic mean.

$\boxed{ \rm{ \:AM \geqslant GM \geqslant HM \: }}$

$\boxed{ \rm{ \: {GM}^{2} = AM \times HM \: }}$