Proving: 2x2+x≤ln⁡(1+x)≤x22+x1+x,x&gt;0.

Hint : Consider f(x)=ln⁡(1+x)−2x2+x and g(x)=x(x+2)2(x+1)−ln⁡(1+x) Show f and g are monotone increasing functions.

I have a shorter proof: by the AM-GM inequality, ∀t&gt;0,11+t≥1(1+t2)2 as well as: ∀t&gt;0,11+t≤1+(1+t)22(1+t)2 Now it is enough to integrate both sides of (1) and (2) over (0,x) to prove the statement.

Let f(x)=2xx+2,g(x)=ln⁡(1+x),h(x)=x2+2x2x+2f(0)=0,g(0)=0,h(0)=0.f′=2(x+2)−2x(x+2)2=4(x+2)2,g′=11+x,h′=(2x+2)(2x+2)−2(x2+2x)(2x+2)2=2x2+4x+4(2x+2)2Now it is easy to check thatf′=4(x+2)2≤g′=11+x≤h′=2x2+4x+4(2x+2)2

Hint : Consider f(x)=ln⁡(1+x)−2x2+x and g(x)=x(x+2)2(x+1)−ln⁡(1+x) Show f and g are monotone increasing functions.

I have a shorter proof: by the AM-GM inequality, ∀t&gt;0,11+t≥1(1+t2)2 as well as: ∀t&gt;0,11+t≤1+(1+t)22(1+t)2 Now it is enough to integrate both sides of (1) and (2) over (0,x) to prove the statement.

Let f(x)=2xx+2,g(x)=ln⁡(1+x),h(x)=x2+2x2x+2f(0)=0,g(0)=0,h(0)=0.f′=2(x+2)−2x(x+2)2=4(x+2)2,g′=11+x,h′=(2x+2)(2x+2)−2(x2+2x)(2x+2)2=2x2+4x+4(2x+2)2Now it is easy to check thatf′=4(x+2)2≤g′=11+x≤h′=2x2+4x+4(2x+2)2

Question: "Proving: 2x2+x≤ln⁡(1+x)≤x22+x1+x,x>0."

feminizmiki

Answered question

2022-01-19

Proving:

\frac{2 x}{2 + x} \leq \ln (1 + x) \leq \frac{x}{2} \frac{2 + x}{1 + x}, x > 0 .

Answer & Explanation

David Clayton

Beginner2022-01-19Added 36 answers

Hint :
Consider

f (x) = \ln (1 + x) - \frac{2 x}{2 + x}

and

g (x) = \frac{x (x + 2)}{2 (x + 1)} - \ln (1 + x)

Show

f

and

g

are monotone increasing functions.

chumants6g

Beginner2022-01-20Added 33 answers

I have a shorter proof: by the AM-GM inequality,

\forall t > 0, \frac{1}{1 + t} \geq \frac{1}{{(1 + \frac{t}{2})}^{2}}

as well as:

\forall t > 0, \frac{1}{1 + t} \leq \frac{1 + {(1 + t)}^{2}}{2 {(1 + t)}^{2}}

Now it is enough to integrate both sides of

(1)

and

(2)

over

(0, x)

to prove the statement.

RizerMix

Expert2022-01-27Added 656 answers

Let

f (x) = \frac{2 x}{x + 2}, g (x) = \ln (1 + x), h (x) = \frac{x^{2} + 2 x}{2 x + 2}

f (0) = 0, g (0) = 0, h (0) = 0.

f^{'} = \frac{2 (x + 2) - 2 x}{(x + 2)^{2}} = \frac{4}{(x + 2)^{2}}, g^{'} = \frac{1}{1 + x}, h^{'} = \frac{(2 x + 2) (2 x + 2) - 2 (x^{2} + 2 x)}{(2 x + 2)^{2}} = \frac{2 x^{2} + 4 x + 4}{(2 x + 2)^{2}}

Now it is easy to check that

f^{'} = \frac{4}{(x + 2)^{2}} \leq g^{'} = \frac{1}{1 + x} \leq h^{'} = \frac{2 x^{2} + 4 x + 4}{(2 x + 2)^{2}}

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