Question

cho 5 số a,b,c,d,e.chứng minh a² +b² +c² +d² +e²  >= e(a+b+c+d)

Answer 1

\(a^2+b^2+c^2+d^2+e^2-e\left(a+b+c+d\right)\)

\(=\left(a^2-ae+\frac{1}{4}e^2\right)+\left(b^2-be+\frac{1}{4}e^2\right)+\left(c^2-ce+\frac{1}{4}e^2\right)+\left(d^2-de+\frac{1}{4}e^2\right)\)

\(=\left(a-\frac{e}{2}\right)^2+\left(b-\frac{e}{2}\right)^2+\left(c-\frac{e}{2}\right)^2+\left(d-\frac{e}{2}\right)^2\ge0\)

\(\Rightarrow a^2+b^2+c^2+d^2+e^2\ge e\left(a+b+c+d\right)\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=d=\frac{e}{2}\)

Answer 2

\(\frac{e}{2}\)

tk minh nhe

moi nguoi

xin do

Answer 3

ckua tkay ai ao tuog vc nku bn