Question

If a,b, c, d are positive real numbers, then show that $(a+b+c+d)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})\ge 16$. What happens when the numbers are all equal?

Answer 1

By applying the inequality A.M.$\ge$ G.M.successively to the set of numbers a, b,c, d ;
$\frac{1}{a},\frac{1}{b},\frac{1}{c},\frac{1}{d}$,we have
$\frac{a+b+c+d}{4}\ge (abcd{)}^{\frac{1}{4}}$
and
$\frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})}{4}\ge {(\frac{1}{a}.\frac{1}{b}.\frac{1}{c}.\frac{1}{d})}^{\frac{1}{4}}$
By multiplying corresponding sides of the above inequalities we get the result.Since the equality holds in ${}^{\prime }A.M.\ge G.M$.'if all the numbers are equal, therefore if the numbers a, b, c, d are not all equal,the strict inequality holds. The condition a, b, c, d are all unequal is a stronger condition than the condition the numbers are not all equal'. Therefore when the numbers are all unequal, the strict inequality holds.