H
huynhbachkhoa23
Cauchy-Schwarz: $a^2+b^2+c^2+d^2 \ge \dfrac{(a+b+c+d)^2}{4}=1$
Cách khác.
1. Giả sử $a\ge b \ge c \ge d$
Theo Chebysev: $\dfrac{a^2+b^2+c^2+d^2}{4} \ge \dfrac{a+b+c+d}{4}.\dfrac{a+b+c+d}{4}$ hay $a^2+b^2+c^2+d^2 \ge 1$
2. Theo Cauchy: $a^2+\dfrac{1}{4} \ge a \rightarrow \sum a^2 \ge \sum a - 1=1$
3. Giả sử $a \ge b \ge c \ge d$
Theo Trebusep: $a.a+b.b+c.c+d.d \ge \dfrac{1}{4}.(a+b+c+d)(a+b+c+d)=1$
4. $f(x)=x^2$
$\dfrac{df}{dx}=2x; \dfrac{d^2f}{(dx)^2}=2 > 0$
$f(x) \ge f'(\dfrac{1}{2})(x-\dfrac{1}{2})+f(\dfrac{1}{2})$
Hay $\sum a^2 \ge \dfrac{1}{2}(\sum a -2)+1=1$