Question

[Ngoc An Pham]

Với mọi số thực a,b,c,d,e ta có:

a²+b²+c²+d²+e² ≥a.(b+c+d+e)

Answer 1

[Trịnh Seiyuu]

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

\(\Rightarrow a^2+b^2+c^2+d^2+e^2-ab-ac-ad-ae\ge0\)

\(\Rightarrow\left(\dfrac{a^2}{4}-ab+b^2\right)+\left(\dfrac{a^2}{4}-ac+c^2\right)+\left(\dfrac{a^2}{4}-ad+d^2\right)+\left(\dfrac{a^2}{4}-ae+e^2\right)\)

\(\Rightarrow\left(\dfrac{a}{2}-b\right)^2+\left(\dfrac{a}{2}-c\right)^2+\left(\dfrac{a}{2}-d\right)^2+\left(\dfrac{a}{2}-e\right)^2\ge0\) (đúng)

Dấu "=" xảy ra khi: \(\dfrac{a}{2}=b=c=d=e\)

Answer 2

[Nhã Doanh]

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

\(\Leftrightarrow4a^2+4b^2+4c^2+4d^2+4e^2\ge4ab+4ac+4ad+4ae\)

\(\Leftrightarrow4a^2+4b^2+4c^2+4d^2+4e^2-4ab-4ac-4ad-4ae\ge0\)

\(\Leftrightarrow\left(a^2-4ab+4b^2\right)+\left(a^2-4ac+4c^2\right)+\left(a^2-4ad+4d^2\right)+\left(a^2-4ae+4e^2\right)\)

\(\Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+\left(a-2e\right)^2\ge0\) ( luôn đúng)