7. If $a, b, c, d$ are positive numbers, and $a+b+c+d=\sqrt{7960}$, then the minimum value of $a^{2}+b^{2}+c^{2}+d^{2}$ is
7. By the convexity of $y=x^{2}$, we know that $\frac{1}{4}\left(a^{2}+b^{2}+c^{2}+d^{2}\right) \geqslant\left(\frac{a+b+c+d}{4}\right)^{2}$. That is, $a^{2}+b^{2}+c^{2}+d^{2} \geqslant \frac{1}{4}(a+b+c+d)^{2}=\frac{1}{4} \cdot 7960=1990$, when $a=b=c=d=\frac{1}{4} \sqrt{7960}=\frac{1}{2} \sqrt{1990}$, the minimum value 1990 is achieved.
There are two basic ideas for finding the conditional extremum. One is to use the condition for equality in inequalities, the most commonly used in high school is the mean value inequality, while the property of convex functions is the most convenient to use.
