12
$\begingroup$

Here is a question that I need to prove

Prove that for $a, b \geq 0$ $$a^8+b^8\geq a^3b^5+a^5b^3$$

So far I have managed to simplify to $$(a^3-b^3)(a^5-b^5)\geq 0$$

$\endgroup$
  • 11
    $\begingroup$ If both $a, b$ are nonnegative, then $a^3 - b^3$ and $a^5-b^5$ will have the same sign. $\endgroup$ – automaticallyGenerated Oct 3 at 3:48
  • 4
    $\begingroup$ $a^3-b^3$ and $a^5-b^5$ have the same sign regardless of whether $a,b$ are nonnegative; this follows from the fact that $x\mapsto x^{5/3}$ is an increasing function. $\endgroup$ – Greg Martin Oct 3 at 23:13
26
$\begingroup$

If $a \ge b$, then $a^3 \ge b^3$ and $a^5 \ge b^5$, which implies $a^3 - b^3 \ge 0$ and $a^5 - b^5 \ge 0$. Since a positive times a positive is positive (or $0$ times anything is $0$), $$(a^3 - b^3)(a^5 -b^5) \ge 0.$$ If instead $a < b$, then $a^3 < b^3$ and $a^5 < b^5$, which implies $a^3 - b^3 < 0$ and $a^5 - b^5 < 0$. Since a negative times a negative is positive, $$(a^3 - b^3)(a^5 -b^5) > 0.$$ Therefore, for all $a,b \ge 0$, $$(a^3 - b^3)(a^5 -b^5) \ge 0.$$

$\endgroup$
  • $\begingroup$ I think you can just have the two cases, (1) $a \gt b \ge 0$ (without loss of generality, since it doesn't matter which you choose as $a$) and (2) $a = b \ge 0$. $\endgroup$ – shoover Oct 3 at 22:37
6
$\begingroup$

Note that $$(a^3-b^3)(a^5-b^5) = (a-b)(a^2+ab+b^2)(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$$

$$=(a-b)^2(a^2+ab+b^2)(a^4+a^3b+a^2b^2+ab^3+b^4)\ge 0$$

$\endgroup$
6
$\begingroup$

I don't think an expansion is needed at all. Here is an even more trivial proof.

Consider that if $a>b$, $a^3 > b^3$ and $a^5> b^5$ since $a$ and $b$ are positive. So, we have that $$(a^3-b^3)(a^5-b^5) > 0$$ when $a>b\geq 0 \tag{1}$.

Now, if $a\leq b$, $b^5\geq a^5$ and $b^3 \geq a^3$ so we have that $$(b^3-a^3)(b^5-a^5) = (a^3-b^3)(a^5-b^5) \geq 0 \tag{2}$$ when $b\geq a\geq 0$.

$(1)$ and $(2)$ are together necessary and sufficient condition to prove the required inequality, i.e. that $$\boxed{a^8+b^8 \geq a^3b^5 + a^5b^3} ~~~\blacksquare$$

$\endgroup$
  • 1
    $\begingroup$ Wow. I could only expect an answer of this quality from a human that is also not a dog. (+1) $\endgroup$ – clathratus Nov 8 at 4:49
3
$\begingroup$

If $a\geq b$ then $a^3\geq b^3$ and $a^5\geq b^5$ so $(a^3-b^3)(a^5-b^5)\geq 0$. Otherwise $a^3-b^3$ is negative, and so is $a^5-b^5$, so their product is positive, and the inequality is proved.

$\endgroup$
3
$\begingroup$

Keep in mind:$$\left(\vphantom{a^2k}a<b\right)\leftrightarrow\left(a^{2k-1}<b^{2k-1};k\in\mathbb{N}\right)\leftrightarrow\left(a^{2k-1}-b^{2k-1}<0;k\in\mathbb{N}\right)$$ $$\to\text{with k=2}: \left(\vphantom{a^2k}a<b\right)\leftrightarrow\left(a^3-b^3<0\right)$$ $$\phantom{\to}\text{with k=3}: \left(\vphantom{a^2k}a<b\right)\leftrightarrow\left(a^5-b^5<0\right)$$ $$\frac{\hphantom{XXXXXXXXXX}}{\hphantom{XXXXXXXXXX}}$$ $$\left(\vphantom{a^2k}a=b\right)\leftrightarrow\left(a^{2k-1}=b^{2k-1};k\in\mathbb{N}\right)\leftrightarrow\left(a^{2k-1}-b^{2k-1}=0;k\in\mathbb{N}\right)$$ $$\to\text{with k=2}: \left(\vphantom{a^2k}a=b\right)\leftrightarrow\left(a^3-b^3=0\right)$$ $$\phantom{\to}\text{with k=3}: \left(\vphantom{a^2k}a=b\right)\leftrightarrow\left(a^5-b^5=0\right)$$ $$\frac{\hphantom{XXXXXXXXXX}}{\hphantom{XXXXXXXXXX}}$$ $$\left(\vphantom{a^2k}a>b\right)\leftrightarrow\left(a^{2k-1}>b^{2k-1};k\in\mathbb{N}\right)\leftrightarrow\left(a^{2k-1}-b^{2k-1}>0;k\in\mathbb{N}\right)$$ $$\to\text{with k=2}: \left(\vphantom{a^2k}a>b\right)\leftrightarrow\left(a^3-b^3>0\right)$$ $$\phantom{\to}\text{with k=3}: \left(\vphantom{a^2k}a>b\right)\leftrightarrow\left(a^5-b^5>0\right)$$

Outgoing from $\left((a^3-b^3)(a^5-b^5)\geq 0\right);\left(\vphantom{b^3}a,b\geq0\right)$:

In case  $a<b$  each of the two factors of the product on the left will be negative, thus the product will be positive.
In case  $a=b$  you get  $0\geq0$ .
In case  $a>b$  each of the two factors of the product on the left will be positive, thus the product will be positive.


I am still suffering an attack of weird humor. Therefore I do this piece of homework by means of -eh- very basic knowledge. October 5, 2019, 02:21:28 UTC +0200: I just did an edit. Before the edit I did not clarify that the subject of the study is not a statement, but a statement pattern.

Prove that for $a,b\geq0$ $$a^8+b^8\geq a^3b^5+a^5b^3$$

In other words:

Prove that all statements of the pattern

$$\left(a^8+b^8\geq a^3b^5+a^5b^3\right);\left(a\vphantom{^8},b\geq0\right)$$

are true.

Remarks:

  1.  I assume:  $\left(a\in\mathbb{R}\right);\left(b\in\mathbb{R}\right)$ .
  2.  $\left(k\in\mathbb{R}\right);\left(n\in\mathbb{N}\right)\to(k^n-1)=\left({\displaystyle\sum_{i=1}^{n}{\left(k^{(n-i)}\right)}}\right)\cdot(k-1)$  .

All these statements in any case are also of pattern:

$$\left(d^8+c^8\geq d^3c^5+d^5c^3\right);\left(\vphantom{d^8}d,c\geq0\right)\mathbf{;\left(\vphantom{d^8}d\geq c\right)}$$

(If  $a=b$ , then assign   $d\mathrel{\mathop:}=a=b$  and  $c\mathrel{\mathop:}=a=b\to c=d=a=b$ .
 If  $a>b$ , then assign  $d\mathrel{\mathop:}=a$  and  $c\mathrel{\mathop:}=b$ .
 If  $b>a$ , then assign  $d\mathrel{\mathop:}=b$  and  $c\mathrel{\mathop:}=a$ .  )

  • Let's look at the case  $\mathbf{\left(d=c\right)}$ :
    $$\left(d^8+c^8≥d^3c^5+d^5c^3\right);\left(\vphantom{d^8}d,c\geq0\right);\left(\vphantom{d^8}d\geq c\right)\mathbf{;\left(\vphantom{d^8}d=c\right)}$$ $$\leftrightarrow$$ $$\left(d^8+d^8≥d^3d^5+d^5d^3\right);\left(\vphantom{d^8}d\geq0\right)$$ $$\leftrightarrow$$ $$\left(2d^8≥2d^8\right);\left(\vphantom{d^8}d\geq0\right)$$
    All statements of this pattern are true.
  • Let's look at the case  $\mathbf{\left(d>c\right)}$ :

    • Let's look at the sub-case  $\mathbf{\left(d>c\right);\left(c=0\right)}$ : $$\left(d^8+c^8\geq d^3c^5+d^5c^3\right);\left(\vphantom{d^8}d,c\geq0\right);\left(\vphantom{d^8}d\geq c\right)\mathbf{;\left(\vphantom{d^8}d>c\right);\left(\vphantom{d^8}c=0\right)}$$ $$\leftrightarrow$$ $$\left(d^8\geq0\right);\left(\vphantom{d^8}d>0\right)$$ All statements of this pattern are true.
    • Let's look at the sub-case  $\mathbf{\left(d>c\right);\left(c>0\right)}$ : $$\left(d^8+c^8\geq d^3c^5+d^5c^3\right);\left(\vphantom{d^8}d,c\geq0\right);\left(\vphantom{d^8}d\geq c\right)\mathbf{;\left(\vphantom{d^8}d>c\right);\left(\vphantom{d^8}c>0\right)}$$ $$\leftrightarrow$$ $$\left(d^8+c^8\geq d^3c^5+d^5c^3\right);\left(\vphantom{d^8}d>c>0\right)$$  $\left(d>c>0\right)\to\exists!k:\left(k>1\right);\left(d=(c\cdot k)\right)$ , thus you can substitute  $d=(c\cdot k); \left(k>1\right)$ : $$\leftrightarrow$$ $$\left((c\cdot k)^8+c^8\geq(c\cdot k)^3c^5+(c\cdot k)^5c^3\right);\left(\vphantom{c^8}c>0\right);\left(\vphantom{c^8}k>1\right)$$ $$\leftrightarrow$$ $$\left(c^8(k^8 + 1)\geq c^8k^3 + c^8k^5\right);\left(\vphantom{c^8}c>0\right);\left(\vphantom{c^8}k>1\right)$$ $$\leftrightarrow$$ $$\left(c^8(k^8 + 1)\geq c^8(k^3 + k^5)\right);\left(\vphantom{c^8}c>0\right);\left(\vphantom{c^8}k>1\right)$$ $$\leftrightarrow$$ $$\left(k^8 + 1\geq k^3 + k^5\right);\left(\vphantom{k^8}k>1\right)$$ $$\leftrightarrow$$ $$\left(k^8 + 1 - k^3 - k^5\geq0\right);\left(\vphantom{k^8}k>1\right)$$ $$\leftrightarrow$$ $$\left(k^8 - k^5 - k^3 + 1\geq 0\right);\left(\vphantom{k^8}k>1\right)$$ $$\leftrightarrow$$ $$\left((k^8 - k^5) - (k^3 - 1)\geq 0\right);\left(\vphantom{k^8}k>1\right)$$ $$\leftrightarrow$$ $$\left(k^5(k^3 - 1) - (k^3 - 1)\geq 0\right);\left(\vphantom{k^5}k>1\right)$$ $$\leftrightarrow$$ $$\left((k^5-1)(k^3 - 1)\geq 0\right);\left(\vphantom{k^5}k>1\right)$$ $$\leftrightarrow$$ $$\left((k^4+k^3+k^2+k+1)(k-1)(k^2+k+1)(k-1)\geq 0\right);\left(\vphantom{k^4}k>1\right)$$ $$\leftrightarrow$$ $$\left((k^4+k^3+k^2+k+1)(k^2+k+1)(k-1)^2\geq 0\right);\left(\vphantom{k^4}k>1\right)$$
      For  $k>1$  all factors on the left side yield positive numbers, thus the product on the left side is positive and therefore  $>0$  and therefore  $\geq 0$ . Therefore:

      All statements of this pattern are true.

Summa summarum:

For all cases of the given statement-pattern you can derive equivalent statement-patterns where all statements of the derived patterns are true.

Therefore all statements of the given statement-pattern are true.

$\endgroup$
2
$\begingroup$

WLOG, consider $a\ge b$. So: $$a^3\ge b^3;a^5\ge b^5$$ Use Rearrangement inequality: $$a^8+b^8\ge a^3b^5+b^3a^5.$$

$\endgroup$

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.