Câu hỏi của Cindy Phương

Question

Cho a+b+c+d=1. Chứng minh: ab+bc+ca<$\dfrac{1}{2}$

Lightning Farron · Accepted Answer

Áp dụng BĐT Cauchy-Schwarz ta có:

$\left(1^2+1^2+1^2+1^2\right)\left(a^2+b^2+c^2+d^2\right)\ge\left(a+b+c+d\right)^2$

$\Rightarrow4\left(a^2+b^2+c^2+d^2\right)\ge\left(a+b+c+d\right)^2=1$

$\Rightarrow a^2+b^2+c^2+d^2\ge\dfrac{1}{4}$

Lại có:

$a^2+b^2+c^2+d^2\ge ab+bc+cd+da\forall a,b,c,d$

$\Rightarrow\dfrac{1}{2}>\dfrac{1}{4}\ge ab+bc+ca+da$ (ĐPCM)

Câu trả lời:

Áp dụng BĐT Cauchy-Schwarz ta có:

$\left(1^2+1^2+1^2+1^2\right)\left(a^2+b^2+c^2+d^2\right)\ge\left(a+b+c+d\right)^2$

$\Rightarrow4\left(a^2+b^2+c^2+d^2\right)\ge\left(a+b+c+d\right)^2=1$

$\Rightarrow a^2+b^2+c^2+d^2\ge\dfrac{1}{4}$

Lại có:

$a^2+b^2+c^2+d^2\ge ab+bc+cd+da\forall a,b,c,d$

$\Rightarrow\dfrac{1}{2}>\dfrac{1}{4}\ge ab+bc+ca+da$ (ĐPCM)