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# help is appreciated

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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 $$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2) \ge (ace + bdf)^2$$

May 22, 2022

#1
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(a) By Cauchy-Schwarz inequality,

$$(c^4 + d^4)(e^4 + f^4) \geq ((ce)^2 + (df)^2)^2$$

Then, again, by Cauchy-Schwarz inequality,

$$\begin{array}{rcl} (a^2 + b^2)^2(c^4 + d^4)(e^4 + f^4) &\geq & (a^2 + b^2)^2((ce)^2 + (df)^2)^2\\ &=& \left((a^2 + b^2)\left((ce)^2+ (df)^2\right)\right)^2\\ &\geq& ((ace + bdf)^2)^2\\ &=& (ace + bdf)^4 \end{array}$$

For (b), you can instead prove that $$(a^2 + b^2)^2 (c^2 + d^2)^2 (e^2 + f^2)^2 \geq (ace + bdf)^4$$. The rest follows from the fact that $$(x + y)^2 \geq x^2 + y^2$$ for nonnegative x, y, and part (a). If you can prove this inequality, the result of part (b) immediately follows.

May 22, 2022