Question

Let a, b, c are three real numbers such that a<b<c. f(x) is continuous in [a, c] and differentiable on (a, c). Also, f(x) is strictly increasing in (a, c). Prove that (b-c)f(a)+(c-a) f(b)+(a-b)f(c)<0

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Answer 1

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

Given that,

➢ a < b < c

and

➢ f(x) is continuous in [a, c] and differentiable on (a, c).

So,

It implies f(x) is continuous in [a, b] and differentiable on (a, b).

So, By Lagranges Mean Value Theorem, there exist atleast one real number d belongs to ( a, b ) such that

$\red{\bf :\longmapsto\:f'(d) = \dfrac{f(b) - f(a)}{b - a} - - - (1)}$

Also,

➢ f(x) is continuous in [a, c] and differentiable on (a, c).

So,

It implies f(x) is continuous in [b, c] and differentiable on (b, c).

So, By Lagranges Mean Value Theorem, there exist atleast one real number e belongs to ( b, c ) such that

$\red{\bf :\longmapsto\:f'(e) = \dfrac{f(c) - f(b)}{c - b} - - - (2)}$

Now,

➢ We have a < d < b < e < c

➢ Further it is given that, f(x) is strictly increasing in (a, c)

So, by definition of strictly increasing function,

$\red{\rm :\longmapsto\:d < e}$

$\bf\implies \:f'(d) < f'(e)$

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

$\rm :\longmapsto\:\dfrac{f(b) - f(a)}{b - a} < \dfrac{f(c) - f(b)}{c - b}$

$\rm :\longmapsto\:f(b)(c - b) - f(a)(c - b) = f(c)(b - a) - f(b)(b - a)$

$\rm :\longmapsto\:f(b)(c - b) - f(a)(c - b) - f(c)(b - a) + f(b)(b - a) < 0$

$\rm :\longmapsto\:f(b)(c - b + b - a) + f(a)( b - c) + f(c)(a - b)< 0$

$\rm :\longmapsto\:f(b)(c- a) + f(a)( b - c) + f(c)(a - b)< 0$

$\rm :\longmapsto\:f(a)( b - c) + f(b)(c - a) + f(c)(a - b)< 0$

can be rewritten as

$\rm :\longmapsto\:(b-c)f(a)+(c-a) f(b)+(a-b)f(c)<0$

Hence, Proved