Question

If a, b, c are positive and in Harmonic progression, then $a^x$ + $c^x$ < 2$b^x$, where x ∈ (0, 1)

• A. True
• B. False

Answer 1

The correct option is B

False

Arithmetic mean of $x^{th}$ powers > $x^{th}$ power of arithmetic mean

$\frac{a^x+c^x}{2}$ > ${\left(\frac{a+c}{2}\right)}^x$

$a^x$ + $c^x$> 2 ${\left(\frac{a+c}{2}\right)}^x$ ---------------(1)

But we know that for 2 numbers a and c, AM > HM

$\frac{a+c}{2}$ > b (b is the HM of a &c because a,b,c are in HP)

${\left(\frac{a+c}{2}\right)}^x$ > $b^x$ ----------------(2)

From (1) & (2),

$a^x$ + $c^x$ < 2$b^x$

So,given statement is False.