$$ \sqrt{1+6x}\le(1+x)^3 $$

I understood the AM is always greater or equal to the GM thing but it was for all values greater than $0$. This doesnt look like its going to be solved the same way. Do we use Cauchy-Schwarz or the AM-GM for this? Don't forget the title for the value of $x$.

  • 1
    $\begingroup$ square both sides of your inequation $\endgroup$ – Dr. Sonnhard Graubner Oct 31 '14 at 14:45

Bernoulli's inequality: Let $t\ge -1$ be a real number, and $n$ a non-negative integer. Then $$ (1+t)^n\ge 1+nt. $$

Let's use $1+3x$ as a stepping stone.

Bernoulli's inequality ($n=3$, $t=x$) gives $$ (1+x)^3\ge 1+3x $$ for all $x\ge-1$.

Also ($n=2$, $t=3x$) $$ (1+3x)^2\ge 1+6x $$ whenever $3x\ge-1$ or, equivalently, when $x\ge-1/3$.

You can take the square root of the latter inequality, if both of them are non-negative (observe the range of $x$ to that end).

And then you can connect the dots.

  • $\begingroup$ You can also do it in one step: $(1+x)^6\ge1+6x$ and then take square roots :-) $\endgroup$ – Jyrki Lahtonen Nov 8 '14 at 14:04

For $0\leq\alpha\leq 1$ the graph $\gamma$ of the function $b_\alpha(y):=(1+y)^\alpha$ is concave, since $$b_\alpha''(y)=\alpha(\alpha-1)(1+y)^{\alpha-2}\leq0\qquad(y\geq-1)\ .$$ It follows that $\gamma$ is for all $y\geq-1$ below its tangent at $(0,1)\in\gamma$. In this way one obtains Bernoulli's inequality $$(1+y)^\alpha\leq 1+\alpha y\qquad(y\geq-1, \ 0\leq \alpha\leq1)\ .$$

With $y:=6x$ and $\alpha:={1\over6}$ we at once obtain the claim.

  • $\begingroup$ didn' expect an expert answer. Not everyone understands the Bernaolli's inequality. How do you explain this problem to a farmer? @Christian Blatter $\endgroup$ – Sherlock Homies Nov 2 '14 at 12:22

Differentiating $RHS-LHS$ gives


which has $0$ as its only real root, so the minimum of the difference occurs at $x=0$, where it is $0$.

Therefore, the inequality holds wherever both $LHS$ and $RHS$ are defined, which is $x\ge-\frac{1}{6}$.

  • $\begingroup$ can we do it alernately without diffrentiating? just using inequality properties? $\endgroup$ – Sherlock Homies Nov 2 '14 at 12:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.