Condition : \(x\ge0\)

We have :

\(P=\dfrac{x^2+2x+1}{x^2+1}=\dfrac{x^2+1}{x^2+1}+\dfrac{2x}{x^2+1}=1+\dfrac{2x}{x^2+1}\)

We got this : 

Following inequality Cauchy , we have :

x² + 1 \(\ge2\sqrt{x^2.1}=2x\)

=> \(\dfrac{2x}{x^2+1}\le\dfrac{2x}{2x}=1\)

=> \(1+\dfrac{2x}{x^2+1}\le2\)

Hence : \(Max_P=2\)

<=> \(\dfrac{2x}{x^2+1}=1\)

<=> 2x = x² + 1

<=> (x - 1)² = 0

<=> x = 1

So , the anwser is A

No further discussion , we will wait for the answer from the teacher .

If me or you're wrong, they both learn from experience . Ok ???

Kayasari Ryuunosuke,P is always not smaller than 0 since we have :

\(x^2+2x+1=\left(x+1\right)^2\ge0;x^2+1>0\) \(\Rightarrow P\ge0\)

I can't understand the sentence "If we want to find the maximum value of P,so P must be positive".It seems you think every algebraic expression has the positive maximum value,doesn't it ?

Moreover,how can you deduce \(x\ge0\) from \(P>0\) ?

Phan Thanh Tinh : If we want to find the Maximum value of P

So , P must be positive 

Means x is a positive number 

=> x \(\ge0\)

Kayasari Ryuunosuke,there's no condition here since \(x^2+1\ge1>0\)

Cauchy's inequality is also known as the AM-GM inequality