Cauchy-Schwartz Inequality problem

Question

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.

Macavity · Accepted Answer

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

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

Now combine the two to get what you want.

Book Of Flames · Answer

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.

Cauchy-Schwartz Inequality problem

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