21By power mean inequality
(4a+1) + ...+ (4d+1)4 ≥ (√4a+1 + ...+√4d+1 4)2
giving 6> 4√2≥√4a+1 + ...+√4d+1
If a,b,c and d are real positive such that :
a + b + c + d = 1
Show that :
√4a+1 + √4b+1 + √4c+1 + √4d+1 < 6
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3 Answers
21
11Also:
Make the transformation a\rightarrow \frac{\tan^2 a}{4}
So the Inequality remains to prove \sum{\sec a}<6
By Tchebycheff Inequality, \left( \sum{ \sec a} \right)^2<4\left( \sum{ \sec ^2 a} \right)=32
The inequality follows
