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How can I tackle the following inequality :

Prove that $4x-x^4 \leq 3$, where $x$ is any real number.

Can someone point me in the right direction?

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    $\begingroup$ you accepted an elegant answer that doesn't require calculus, but if you know a little calculus, it is easy: let $f(x) = 4x -x^4$. $f(x) \to -\infty$ as $|x| \to \infty$, so $f$ achieves a maximum value at some $x_0$. $f'(x_0) = 0$. $f'(x) = 4 - 4x^3$ so $x_0 = 1$. $f(1) = 3$ is the maximum value of $f$. $\endgroup$
    – Stefan Smith
    Apr 14, 2013 at 22:59

6 Answers 6

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You could also apply $\mathrm{AM}-\mathrm{GM}$

$$\frac{x^{4}+1+1+1}{4} \geq \sqrt[4]{x^{4} 1\cdot 1\cdot 1} =|x|$$

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  • $\begingroup$ Wow! what a nice a nice approach. thanks a lot @clark. +1 from me. $\endgroup$
    – user53386
    Apr 14, 2013 at 9:07
  • $\begingroup$ @user53386 glad I could help:) $\endgroup$
    – clark
    Apr 14, 2013 at 10:33
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We see that , $x^4-4x+3=x^4-2x^2+1+2x^2-4x+2=(x^2-1)^2+2(x-1)^2 \geq 0$ and so $-(x^4-4x+3)\leq 0 \implies -x^4+4x-3 \leq 0$ and hence $4x-x^4 \leq 3$

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Hint: $4x-x^4-3=-(x-1)^2((x+1)^2+2)\le0$

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For your question, consider rearranging the inequality to $$\begin{align*}4x-x^4 &\leq 3\\4x-x^4-3 &\leq 0 \end{align*}$$ Now let's prove it. For all real $x$, consider the stationary value of the function

(let's say I didn't read the question and do not know if the stationary value is a minimum or maximum)

$$f(x)=4x-x^4-3$$ First order condition:$$\begin{align*}f'(x)=4-4x^3&=0\\4x^3&=4\\x^3&=1\\x&=1\end{align*}$$ Second order condition:$$\begin{align*}f''(x)=-12x^2&=-12\end{align*}$$ and so we conclude the maximum value of $f$ is at $x=1$. $f(1)=4-1-3=0$. We say that $f$ can never go beyond $0$, or $f(x)\leq0$ for all real $x$. It agrees with the initial inequality and hence it is true.

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  • $\begingroup$ Thanks a lot @bryansis2010 . Nice approach. +1 from me. $\endgroup$
    – user53386
    Apr 14, 2013 at 8:59
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You can also do this by some calculus. It is clear for $x \leq 0$. $4x - x^4$ is increasing until $x = 1$, where we have $4x - x^4 = 3$ on the dot, so we have it for $x \in [0,1]$. Then it decreases afterwards, so we have it.

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Let $y = 4x-x^4-3$. Note that $y$ has just 1 extrema: $\sqrt[3]{(\frac 4 3)}$. $y$ is not positive at this point, so as isn't at $\infty$ and$-\infty$, therefore it isn't positive at $(-\infty;~\infty)$.

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