Proof of $ x^5 \leq \frac{56}{3} + \frac{5}{3} x^3 \Rightarrow x \leq 2$

Question

I have a problem (but unfortunately no solution to it) where I have to prove that for every $x \in \mathbb{R}$ which satisfies $ x^5 \leq \frac{56}{3} +\frac{5}{3}x^3 $, holds $x \leq 2 $ . So to put it another way: $$ x^5 \leq \frac{56}{3} + \frac{5}{3} x^3 \Rightarrow x \leq 2$$

rewritten this gives

$$ 3x^5 \leq 56 + 5x^3 \Rightarrow x \leq 2$$

How do I have to proceed? An induction works only with integers, for a direct proof I don't have the idea. My proof sketch was this, but it's obvisouly false :/

$$3x^5 \leq 56 + 5x^3 \Rightarrow x \leq 2 \\ x^5 \leq 56 + 5x^3 \\ x^2 \leq 56/x^3 + 5 \\ x^2 - \frac{56}{x^3} \leq 5 $$

Answer 1

\begin{equation*} x^{5}\leq \frac{56}{3}+\frac{5}{3}x^{3} \end{equation*} is equivalent to \begin{equation*} 3x^{5}-5x^{3}-56\leq 0\text{,} \end{equation*} where \begin{equation*} 3x^{5}-5x^{3}-56=\left( 3x^{4}+6x^{3}+7x^{2}+14x+28\right) \times \left( x-2\right) \text{.} \end{equation*} Notice that $3x^{4}+6x^{3}+7x^{2}+14x+28>0$ holds for all $x\in \mathbb{R}$ since we have \begin{equation*} 3x^{4}+6x^{3}+7x^{2}+14x+28=3x^{2}\left( x+1\right) ^{2}+4(x+\frac{7}{4} )^{2}+\frac{63}{4}\text{.} \end{equation*} Thus, $3x^{5}-5x^{3}-56\leq 0$ is satisfied if and only if $x\leq 2$. Hope this helps.

Answer 2

Not using derivatives, you would have (on the left),

$$\forall x\in \mathbb{R}, 3x^5 - 5x^3 \leq 56$$

equivalent to, factorizing by $x^3$,

$$\forall x\in \mathbb{R}, x^3(3x^2-5)\leq 56$$

Now, suppose that $x>2$. Then, you would have $x^3>2^3=8$ and $3x^2-5>3\cdot 2^2 - 5 = 7$ and multiplying both terms (since they are positive) yields $x^3(3x^2-5)>8\times 7=56$, thus a contradiction.

So, $x\leq 2$.