Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$

Question

My try

My book gives as a hint to move everything to the left hand side of the inequality and then factor and see what I get in the long factorization process and to lookout for squares.

So that's what I have tried:

\begin{array} ((a^7+b^7)(a^2+b^2) &\ge (a^5+b^5)(a^4+b^4) \\\\ (a^7+b^7)(a^2+b^2)-(a^5+b^5)(a^4+b^4) &\ge 0 \\\\ (a+b)(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)(a^2+b^2)-(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)(a^4+b^4) &\ge 0 \\\\ (a+b)\left[(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)((a+b)^2-2ab)-(a^4+b^4)(a^4-a^3b+a^2b^2-ab^3+b^4)\right] &\ge 0 \end{array}

Now it's not clear what I have to do next.I am stuck.

Note: My book doesn't teach any advanced technique for solving inequality as AM-GM ,Cauchy inequality etc..

Did you misinterpret like expanding $a^27+b^7$. No i said multiplying brackets. — Archis Welankar, Jan 05 '16 at 09:35

score 4 · Accepted Answer · answered Jan 05 '16 at 09:32

4

You can simplify considerably by multiplying out both sides $$a^9+a^7b^2+a^2b^7+b^9\ge a^9+a^5b^4+a^4b^5+b^9$$

Cancelling the common terms from each side and dividing through by $a^2b^2$ gives $$a^5+b^5\ge a^3b^2+a^2b^3$$

Now move everything to the LHS and factorise.

Note if $a$ or $b$ is zero, equality is obvious.

answered Jan 05 '16 at 09:32

Mark Bennet

100,194

Why can I divide by $a^2b^2$ ?what if $a$ or $b$ is $0$ ? – Mr. Y Jan 05 '16 at 09:40
See my comment at the end - once you have eliminated the possibility that $ab=0$ you have $ab\gt 0$ and can divide through. Of course you can leave that factor in too. But faced with a complex problem, I would always try to simplify as much as possible. – Mark Bennet Jan 05 '16 at 09:42

score 4 · Answer 2 · answered Jan 05 '16 at 09:33

4

\begin{align*} (a^7 + b^7)(a^2 + b^2) - (a^5 + b^5)(a^4 + b^4) &= a^7 b^2 + a^2 b^7 - a^5 b^4 - a^4 b^5 \\ &= a^2 b^2 \big(a^5 + b^5 - a^3 b^2 - a^2 b^3\big) \\ &= a^2 b^2 \big(a^2 (a^3 - b^3) + b^2 (b^3 - a^3)\big) \\ &= a^2 b^2 (a^2 - b^2)(a^3 - b^3) \end{align*}

Now regardless of how $a$ and $b$ are related to each other, the two difference terms above have the same sign.

answered Jan 05 '16 at 09:33

As short and direct as possible! – DanielWainfleet Jan 05 '16 at 23:10

score 4 · Answer 3 · answered Jan 05 '16 at 09:33

Multiplying out, this is equivalent to $a^7b^2 + a^2b^7 \geq a^5b^4 + a^4b^5$, or $a^5 + b^5 \geq a^3 b^2 + a^2b^3$ (the case $ab=0$ is easy).

We can prove this using weighted AM-GM:

$$\frac{3}{5}a^5 + \frac{2}{5}b^5 \geq a^3 b^2$$ $$\frac{2}{5}a^5 + \frac{3}{5}b^5 \geq a^2 b^3$$

(Sorry, I didn't notice that you specified no AM-GM, but I'll leave this up in case others are interested.)

score 3 · Answer 4 · answered Jan 05 '16 at 09:49

We may suppose that $a\geq b>0$. Hence dividing by $b^9$ and putting $x=a/b$, we have to show that for $x\geq 1$, we have $(x^7+1)(x^2+1)\geq (x^5+1)(x^4+1)$ or $$\frac{x^7+1}{x^5+1}\geq \frac{x^4+1}{x^2+1}$$ For $x\geq 1$ fixed, put $\displaystyle f(u)=\frac{x^2u+1}{u+1}$. It is easy to see that $f$ is increasing on $[0,+\infty[$; as $x^5\geq x^2$, we get $f(x^5)\geq f(x^2)$ and we are done.

Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$

4 Answers4