# Cauchy-Schwartz Inequality problem

Mathematics Asked by In-finite on December 10, 2020

Let $$a,$$ $$b,$$ $$c,$$ $$d,$$ $$e,$$ $$f$$ be nonnegative real numbers.

(a) Prove that
$$(a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) ge (ace + bdf)^4.$$

(b) Prove that
$$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2) ge (ace + bdf)^2.$$

I’m not sure how I should start approaching both problems. I believe I should use Cauchy-Schwartz, but I’m not sure.
Any help would be appreciated! Thanks in advance.

These are both CS inequality applications, you should try yourself. Here is the first one:

1. $$(c^4+d^4)(e^4+f^4) geqslant (c^2e^2+d^2f^2)^2$$
2. $$(a^2+b^2)(c^2e^2+d^2f^2) geqslant (ace+bdf)^2$$

Now combine the two to get what you want.

Correct answer by Macavity on December 10, 2020

For (a),
Just use Holder's Inequality.

For (b),
Expand $$(a^2 + b^2) (c^2 + d^2) (e^2 + f^2)$$. (It is equivalent to the LHS of my answer)
Expand $$(ace + bdf)^2$$. (It is equivalent to the RHS of my answer)
The problem is just proving $$a^2 c^2 e^2 + a^2 c^2 f^2 + a^2 d^2 e^2 + a^2 d^2 f^2 + b^2 c^2 e^2 + b^2 c^2 f^2 + b^2 d^2 e^2 + b^2 d^2 f^2 ge a^2 c^2 e^2 + b^2 d^2 f^2 + 2abcdef$$ or $$a^2 c^2 f^2 + a^2 d^2 e^2 + a^2 d^2 f^2 + b^2 c^2 e^2 + b^2 c^2 f^2 + b^2 d^2 e^2 ge 2abcdef quad textrm{By cancelling 2 terms on the left and right side.}$$ Now, that is just $$(acf - bde)^2 + a^2 d^2 e^2 + a^2 d^2 f^2 + b^2 c^2 e^2 + b^2 c^2 f^2 ge 0$$ We have proved by this that the inequality is valid for any real numbers, not only non-negative ones.

Answered by Book Of Flames on December 10, 2020

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