1. Prove that $a^4 + b^4 > 2a^2 b^2$ for any real numbers $a, b$. Start with the identity for squares: $$ (a^2 - b^2)^2 \geq 0 $$ Expanding: $$ a^4 - 2a^2 b^2 + b^4 \geq 0 $$ Rearranging: $$ a^4 + b^4 \geq 2a^2 b^2 $$ Since squares are non-negative and equality holds only if $a^2 = b^2$, for strict inequality $a^4 + b^4 > 2a^2 b^2$ when $a^2 \neq b^2$. 2. Prove that $a^4 + b^4 + c^4 + d^4 \geq 4abcd$ for any real numbers $a, b, c, d$. Use the AM-GM inequality (Arithmetic Mean - Geometric Mean): $$ \frac{a^4 + b^4 + c^4 + d^4}{4} \geq \sqrt[4]{a^4 b^4 c^4 d^4} $$ Simplify the right side: $$ \sqrt[4]{a^4 b^4 c^4 d^4} = |abcd| $$ Multiply both sides by 4: $$ a^4 + b^4 + c^4 + d^4 \geq 4|abcd| $$ Since $4abcd \leq 4|abcd|$, it follows that: $$ a^4 + b^4 + c^4 + d^4 \geq 4abcd $$ Final answers: 1. $a^4 + b^4 > 2a^2 b^2$ (strict inequality if $a^2 \neq b^2$) 2. $a^4 + b^4 + c^4 + d^4 \geq 4abcd$