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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 $$(a^3-b^3)(a^5-b^5)\geq 0$$

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    $\begingroup$ If both $a, b$ are nonnegative, then $a^3 - b^3$ and $a^5-b^5$ will have the same sign. $\endgroup$ Oct 3, 2019 at 3:48
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    $\begingroup$ $a^3-b^3$ 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. $\endgroup$ Oct 3, 2019 at 23:13

6 Answers 6

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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 $a,b \ge 0$, $$(a^3 - b^3)(a^5 -b^5) \ge 0.$$

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  • $\begingroup$ 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, 2019 at 22:37
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Note that $$(a^3-b^3)(a^5-b^5) = (a-b)(a^2+ab+b^2)(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$$

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

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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 $$\boxed{a^8+b^8 \geq a^3b^5 + a^5b^3} ~~~\blacksquare$$

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    $\begingroup$ Wow. I could only expect an answer of this quality from a human that is also not a dog. (+1) $\endgroup$
    – clathratus
    Nov 8, 2019 at 4:49
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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.

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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.

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

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