1. **Problem:** Show that for any real numbers $a$ and $b$, the inequality $a^4 + b^4 > 2a^2b^2$ holds. 2. **Formula and Explanation:** We use the fact that the square of any real number is non-negative: for any real $x$, $x^2 \geq 0$. Also, the inequality resembles the form of the square of a difference. 3. **Step-by-step proof:** - Consider the expression $(a^2 - b^2)^2$. - Expanding it, we get: $$ (a^2 - b^2)^2 = a^4 - 2a^2b^2 + b^4 $$ - Since squares are always non-negative, we have: $$ (a^2 - b^2)^2 \geq 0 $$ - Rearranging gives: $$ a^4 + b^4 \geq 2a^2b^2 $$ - The equality holds if and only if $a^2 = b^2$, otherwise the inequality is strict. 4. **Conclusion:** Thus, for any real numbers $a$ and $b$, $$ a^4 + b^4 \geq 2a^2b^2 $$ with equality if $a^2 = b^2$. --- 1. **Problem:** Show that for any real numbers $a, b, c, d$, the inequality $$ a^4 + b^4 + c^4 + d^4 \geq 4abcd $$ holds. 2. **Formula and Explanation:** This is an application of the Arithmetic Mean - Geometric Mean (AM-GM) inequality, which states that for non-negative real numbers $x_1, x_2, ..., x_n$, $$ \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n} $$ with equality if and only if all $x_i$ are equal. 3. **Step-by-step proof:** - Since $a^4, b^4, c^4, d^4$ are all non-negative (even powers), apply AM-GM to these four numbers: $$ \frac{a^4 + b^4 + c^4 + d^4}{4} \geq \sqrt[4]{a^4 b^4 c^4 d^4} $$ - 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 $4abcd \leq 4|abcd|$, the inequality $$ a^4 + b^4 + c^4 + d^4 \geq 4abcd $$ holds for all real $a,b,c,d$. 4. **Conclusion:** The inequality is true for all real numbers $a,b,c,d$, with equality if and only if $a^4 = b^4 = c^4 = d^4$ and $abcd \geq 0$.