# Prove the inequality is true

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

• If both $a, b$ are nonnegative, then $a^3 - b^3$ and $a^5-b^5$ will have the same sign. – automaticallyGenerated Oct 3 '19 at 3:48
• $a^3-b^3$ 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. – Greg Martin Oct 3 '19 at 23:13

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

• 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) $a = b \ge 0$. – shoover Oct 3 '19 at 22:37

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

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

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

• Wow. I could only expect an answer of this quality from a human that is also not a dog. (+1) – clathratus Nov 8 '19 at 4:49

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.

Keep in mind:$$\left(\vphantom{a^2k}a $$\to\text{with k=2}: \left(\vphantom{a^2k}a $$\phantom{\to}\text{with k=3}: \left(\vphantom{a^2k}a $$\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  $$a  each of the two factors of the product on the left will be negative, thus the product will be positive.
In case  $$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.

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