Question

[Hanh Do]

CMR \(\sqrt{\left(a-b\right)^2+c^2}+\sqrt{\left(a+b\right)^2+c^2}\ge2\sqrt{a^2+c^2}\)

Answer 1

[Trần Thanh Phương]

Áp dụng bất đẳng thức Mincopxki :

\(\sqrt{\left(a-b\right)^2+c^2}+\sqrt{\left(a+b\right)^2+c^2}\ge\sqrt{\left[\left(a-b\right)+\left(a+b\right)\right]^2+\left(c+c\right)^2}\)

\(=\sqrt{\left(2a\right)^2+\left(2c\right)^2}\)

\(=\sqrt{4a^2+4c^2}\)

\(=2\sqrt{a^2+c^2}\) ( đpcm )

Dấu "=" xảy ra \(\Leftrightarrow\frac{a-b}{c}=\frac{a+b}{c}\Leftrightarrow a-b=a+b\Leftrightarrow-b=b\Leftrightarrow b=0\)

*) Chứng minh bđt Mincopxki cho 2 bộ số :

\(\sqrt{x^2+y^2}+\sqrt{z^2+t^2}\ge\sqrt{\left(x+z\right)^2+\left(y+t\right)^2}\)

\(\Leftrightarrow x^2+y^2+z^2+t^2+2\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}\ge\left(x+z\right)^2+\left(y+t\right)^2\)

\(\Leftrightarrow x^2+y^2+z^2+t^2+2\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}\ge x^2+y^2+z^2+t^2+2xz+2yt\)

\(\Leftrightarrow\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}\ge xz+yt\)

\(\Leftrightarrow\left(x^2+y^2\right)\left(z^2+t^2\right)=\left(xz+yt\right)^2\)

\(\Leftrightarrow x^2z^2+x^2t^2+y^2z^2+y^2t^2\ge x^2z^2+2xyzt+y^2t^2\)

\(\Leftrightarrow x^2t^2-2xyzt+y^2z^2\ge0\)

\(\Leftrightarrow\left(xt-yz\right)^2\ge0\) ( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow xt=yz\Leftrightarrow\frac{x}{y}=\frac{z}{t}\)

Answer 2

[Trần Thanh Phương]

Dòng đầu viết lỗi rồi, ai tốt bụng sửa hộ tớ :

\(...\ge\sqrt{\left[\left(a-b\right)+\left(a+b\right)\right]^2+\left(a+c\right)^2}\) nha :)