Prove that if $$a^3-b^3=2 \ \text{ and }\ a^5-b^5 \ge 4$$ then $$a^2+b^2 \ge 2.$$
If $a^3-b^3=2 \ \text{ and }\ a^5-b^5 \ge 4$, then $a^2+b^2 \ge 2$
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algebra-precalculus
inequality
polynomials
factoring
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0what kind of numbers are $$a,b$$? – 2017-01-08
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4Maybe start with $$(a^3-b^3)(a^2+b^2)=a^5-b^5+a^2 b^2(a-b)$$ – 2017-01-08
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0a, b can be any kind of numbers. If we start with the thing which is above, we will have to prove that a^2b^2(a-b) is always positive, then we will solve it. But how to do it? – 2017-01-08
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0$$a^3-b^3=(a-b)(a^2+b^2+ab)=2$$ You need to show that $a^2+b^2+ab>0$ – 2017-01-08
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0@idliketodothis: $a^3 - b^3 = 2$ implies that $a > b$ ... – 2017-01-08
1 Answers
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From the conditions we have $\frac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^2+ab+b^2}\geq2$.
Thus, it remains to prove that $a^2+b^2\geq\frac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^2+ab+b^2}$, which is $a^2b^2\geq0$.
Done!
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0How does proving that a^2+b^2 is greater or equal than the fraction help in solving the problem? – 2017-01-08
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0@idliketodothis If $a^2+b^2\geq\frac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^2+ab+b^2}$ and $\frac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^2+ab+b^2}\geq2$ so $a^2+b^2\geq2$. – 2017-01-08