Prove if $x > 0$, $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$
Prove if $x > 0$, $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$.
I find online, one person suggested using Taylor Theorem to expand the right-hand side, and apply Bernoulli's inequality.
So, if $x_0 = 0$, $\sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{4(2!)}x^2+R$, where R is larger than $0$, this makes sense to me, but I'm trying to find another way to prove the equality. Like Mean Value Theorem for inequality $\sqrt{1+x} \le 1+\frac{1}{2}x$.
I see the part where $1+\frac{1}{2}x$ is the same, but get trouble to put $\frac{1}{8}x^2$ into the equation.
$\endgroup$ 23 Answers
$\begingroup$We’ll let $a = x + 1$
Transform the inequality by substitute $x$ with $a - 1$.
After some calculations we the get
$$ -a^2 + 6a + 3 \leq 8\sqrt{a} $$
Square both sides and we get
$$ a^4 - 12a^3 + 30a^2 - 28a + 9 \leq 0$$
Which is just $(a-1)^3(a - 9) \leq 0$.
Since $a > 1$ , this statement is true for all $a \in [1,9]$.
Thus, we are left to prove this statement for all $a > 9$.
Since $a = x + 1$ , $x > 8$.
Consider $(x - 2)^2 + 12 > 0$
After some calculations we then get $\frac{1}{2}x < 3 - 1 + \frac{1}{8}x^2$.
But since $x > 8$ we then get $1 + \frac{1}{2}x - \frac{1}{8}x^2 < \sqrt{x + 1}$.
Hence, this inequality is true for all $x > 0$.
$\endgroup$ 0 $\begingroup$The inequality can be proved using elementary algebra. If $\dfrac{1}{8}x^2 < 1+\dfrac{1}{2}x\implies 2-2\sqrt{3} <x < 2+2\sqrt{3} < 8$, then both sides are positive, and square both of them: $\left(1+\dfrac{1}{2}x-\dfrac{1}{8}x^2\right)^2 < 1+x\iff \left(1+\dfrac{1}{2}x\right)^2-2\left(1+\dfrac{1}{2}x\right)\left(\dfrac{1}{8}x^2\right)+\left(\dfrac{1}{8}x^2\right)^2< 1+x\iff -x+\dfrac{1}{8}x^2 < 0\iff \dfrac{1}{8}x^2 < x\iff x^2 - 8x < 0\iff x(x-8) < 0\iff 0 < x < 8$, and this is true since we've shown that $8 > x > 0$, and if $\dfrac{1}{8}x^2 \ge 1+\dfrac{1}{2}x$, then it is obviously true since the left side is non-positive and the right side is non-negative. You're done !
$\endgroup$ $\begingroup$Well putting $x=\sinh^2(a)$
We have to show (using $\cosh^2(a)-\sinh^2(a)=1$):
$$\cosh(a)\geq 1+\frac{1}{2}\sinh^2(a)-\frac{1}{8}\sinh^4(a)$$
Or :
$$\sinh^6\Big(\frac{a}{2}\Big)(\cosh(a)+3)\geq 0$$
Wich is clearly obvious !
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