\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow ab+bc+ca=0\)

\(a+b+c=1\Rightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=1\)

\(\Rightarrow a^2+b^2+c^2=1\)

dung la may dua hoc ngu nhu bo

Đề: Cho a, b, c, d là 4 số dương thoả mãn abcd = 1. Chứng minh rằng: \(\left(\sqrt{1+a}+\sqrt{1+b}\right)\left(\sqrt{1+c}+\sqrt{1+d}\right)\ge8\)

~ ~ ~ ~ ~

Áp dụng BĐT AM - GM, ta có:

\(\left(\sqrt{1+a}+\sqrt{1+b}\right)\left(\sqrt{1+c}+\sqrt{1+d}\right)\)

\(\ge2\sqrt[4]{\left(1+a\right)\left(1+b\right)}\times2\sqrt[4]{\left(1+c\right)\left(1+d\right)}\)

\(=4\sqrt[4]{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\)

\(\ge4\sqrt[4]{2\sqrt{a}\times2\sqrt{b}\times2\sqrt{c}\times2\sqrt{d}}\)

\(=4\sqrt[4]{16\sqrt{abcd}}\)

= 8 (đpcm)

Dấu "=" xảy ra khi a = b = c = d = 1

\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)

\(=\sqrt{\left(1-3x\right)^2}+\sqrt{\left(3x-2\right)^2}\)

\(=\left|1-3x\right|+\left|3x-2\right|\)

\(\ge\left|1-3x+3x-2\right|=\left|-1\right|=1\)

Dấu "=" xảy ra \(\Leftrightarrow\left(1-3x\right)\left(3x-2\right)\ge0\Leftrightarrow\frac{1}{3}\le x\le\frac{2}{3}\)

Vậy \(A_{min}=1\) tại \(\frac{1}{3}\le x\le\frac{2}{3}\)

Xin lỗi cậu tớ mới học lớp 7 thôi

Đặt x+1/x=t thì x^2+1/x^2=t^2-2

Phương trình viết lại:3(t^2-2)-16t(+-)26=0

Sau đó quy về giải phương trình bậc hai là được

\(P=x^2-x\sqrt{y}+x+y-\sqrt{y}+1\)

\(\Leftrightarrow2P=2x^2-2x\sqrt{y}+2x+2y-2\sqrt{y}+2\)

\(=\left[\left(x^2-2x\sqrt{y}+y\right)+\frac{4}{3}.\left(x-\sqrt{y}\right)+\frac{4}{9}\right]+\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)+\left(y-\frac{2}{3}.\sqrt{y}+\frac{1}{9}\right)+\frac{4}{3}\)

\(=\left(x-\sqrt{y}+\frac{2}{3}\right)^2+\left(x+\frac{1}{3}\right)^2+\left(\sqrt{y}-\frac{1}{3}\right)^2+\frac{4}{3}\ge\frac{4}{3}\)

\(\Rightarrow P\ge\frac{2}{3}\)