solution verification| Determine the smallest number that is integer $k$ for which the inequality $x^4+4x^2+k>6x$ is true for every real number x.

Question

the question

Determine the smallest number integer $k$ for which the inequality $x^4+4x^2+k>6x$ is true for every real number $x$.

my idea

$x^4+4x^2-6x+4>4-k => x^4+4x^2-6x+4=(x^2-2x+1)(x^2+2x+4)+3x^2=(x-1)^2((x+1)^2+3)+3x^2 \geq 3$

This means that $4-k<3=> -k<-1=> k>1$ so the smallest value it can take is 2.

Im not sure is this inequality $(x-1)^2((x+1)^2+3)+3x^2 \geq 3$ is right. I don't know how to show it and I conclude it by doing brute force. I hope one of you can help me and tell me how is the correct way to solve it! Thank you.

Answer 1

Im not sure is this inequality $(x-1)^2((x+1)^2+3)+3x^2 \geq 3$ is right

It is not right. For $x=\frac 12$, the LHS is equal to $\frac{33}{16}$ which is smaller than $3$.

We can show $$(x-1)^2\bigg((x+1)^2+3\bigg)+3x^2 \geq \frac 32$$ as follows :

$$\begin{align}&(x-1)^2\underbrace{\bigg((x+1)^2+3\bigg)}_{\ge 3}+3x^2 \\\\&\ge (x-1)^2\times 3+3x^2 \\\\&=6x^2-6x+3 \\\\&=6\bigg(x-\frac 12\bigg)^2+\frac 32 \\\\&\ge\frac 32\end{align}$$

So, for every real number $x$, $$x^4+4x^2-6x\ge \frac 32-4=-\frac 52$$ holds.

So, if $k\ge 3$, then $$x^4+4x^2-6x\ge -\frac 52\gt -k$$ holds.

This means we can say that if $k\ge 3$, then for every real number $x$, the inequality $x^4+4x^2+k\gt 6x$ holds.

On the other hand, $k\ge 3$ is necessary since taking $x=\frac 35$, we have $$k\gt 6x-x^4-4x^2=\frac{1269}{625}\gt 2$$

In conclusion, under the condition that $k$ is an integer, we can say that for every real number $x$, the inequality $x^4+4x^2+k\gt 6x$ holds if and only if $k\ge 3$.

Therefore, the smallest such integer is $\color{red}3$.

Answer 2

Your answer is wrong. Instead, note that $k$ must be positive (set $x=0$ for this), and the inequality is obvious for $x\le0$. Henceforth, $x> 0$. We use AM_GM:

$$\frac{x^4+4x^2+4\cdot (k/4)}{6} \ge x\sqrt[6]{\left(\frac{k}{4}\right)^4 \cdot 4}$$ Intuitively, it seems if we have $\left(\frac{k}{4}\right)^4 \cdot 4 \ge 1$, then we'll be good, provided we show no smaller $k$ works. We observe that this inequality is false for $k=2$ and true for $k=3$.

Now, it remains to be shown $k=2$ does not work in the original question. Just put $x=0.6$ for this which gives $3.5696>3.6$. Finally:

$$\frac{x^4+4x^2+4\cdot (3/4)}{6} \ge x\sqrt[6]{\left(\frac{3}{4}\right)^4 \cdot 4} > x$$ Hence$$\boxed{x^4+4x^2+3 > 6x}$$

Answer 3

Consider the polynomial $P(x)=x^4+4x^2-6x$, we have $P'(x)=4x^3+8x-6$ and $P''(x)=12x^2+8$. As $P''$ is always stricly positive we conclude that $P$ is convex, in paritcular, its graph lies above any of its tangent. One computes, $P'(1/2)=-3/2$, $P(1/2)=1/16-2$, $P'(1)=6$ and $P(1)=-1$.

Hence $P(x)\geq -3/2(x-1/2)+1/16-2$, In partcular we have that $\forall x\in(-\infty,1], P(x)\geq -3/2(1-1/2)+1/16-2=-(11/16+2)>-3$.

Simillarly, $\forall x\in[1,\infty),P(x)\geq 6(x-1)-1\geq -1>-3$.

Therefore we have that $\forall x\in\mathbb{R},P(x)>-3$. As aforementionned $k=2$ does not work.

Answer 4

Another way.

$$f(x)=-x^4-4x^2+6x$$ $$f'(x)=-4x^3-8x+6$$ The maximum of $f(x)$ occurs at $(x_m,y_m)$ where $f'(x_m)=0$. By intermediate-value theorem, $\tfrac35<x_m<\tfrac45.$ Hence $$f(\tfrac35)=\tfrac{1269}{625}<y_m<f(\tfrac35)+f'(\tfrac35)(\tfrac45-\tfrac35)=\tfrac{1311}{625}<k.$$ Conclusion: The smallest such integer $k$ is $k=3$.