Ta có BĐT \(x^2+1\ge2x\Leftrightarrow\left(x-1\right)^2\ge0\)

Tương tự cũng có 2 BĐT tương tự:

\(y^2+1\ge2y;z^2+1\ge2z\)

\(\Rightarrow x^2+y^2+z^2+3\ge2\left(x+y+z\right)\left(1\right)\)

Và BĐT \(x^2+y^2+z^2\ge xy+yz+xz\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(y-z\right)^2\ge0\)

\(\Rightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\left(2\right)\)

Cộng theo vế 2 BĐT (1) và (2) có:

\(3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+xz\right)\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)+3\ge2\cdot6=12\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge9\Leftrightarrow x^2+y^2+z^2\ge3\)

Xảy ra khi \(x=y=z=1\)

Lớp 9 gì mà hs lớp 7 làm đc :)) ahaha

Áp dụng bất đẳng thức Cauchy ta có :

\(x^2+1\ge2x\)

\(y^2+1\ge2y\)

\(z^2+1\ge2z\)

\(x^2+y^2\ge2xy\)

\(y^2+z^2\ge2yz\)

\(x^2+z^2\ge2zx\)

Cộng vế với vế ta được :

\(3x^2+3y^2+3z^2+3\ge x+y+z+xy+xz+yz\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)+3\ge6\)

\(\Rightarrow x^2+y^2+z^2\ge\frac{6-3}{3}=1\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)

Vậy \(x^2+y^2+z^2\) có GTNN là 1 tại \(x=y=z=1\)

Có ab > 2013a + 2014b <=> 1 > 2013/b + 2014/a (vì a,b >0 )

\(\Leftrightarrow a+b>\frac{2013\left(a+b\right)}{b}+\frac{2014\left(a+b\right)}{a}=2013+2014+\frac{2013a}{b}+\frac{2014b}{a}\)

Mà \(\frac{2013a}{b}+\frac{2014b}{a}\ge2\sqrt{2013\cdot2014}\)

\(\Rightarrow a+b>2013+2014+2\sqrt{2013\cdot2014}=\left(\sqrt{2013}+\sqrt{2014}\right)^2\)

=> đpcm

Tích cho mk nhoa !!!! ~~~

\(4\sqrt[4]{a}+7\sqrt[7]{b}\ge11\sqrt[11]{ab}\)

Ta có: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)

\(\Rightarrow\frac{a^3+b^3+c^3}{4abc}=\frac{3}{4}+\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{4abc}\)

\(=\frac{3}{4}+\frac{1}{4}\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(\ge\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{3}{2}\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\ge\frac{9}{ab+ac+bc}\right)\)

\(\Rightarrow\frac{a^3+b^3+c^3}{4abc}\ge\frac{9}{4}\left(\frac{a^2+b^2+c^2}{ab+bc+ac}\right)-\frac{3}{2}\left(1\right)\)

Lại có:\(\frac{\left(a+b+c\right)^2}{30\left(a^2+b^2+c^2\right)}=\frac{a^2+b^2+c^2+2\left(ab+bc+ac\right)}{30\left(a^2+b^2+c^2\right)}\)

\(=\frac{1}{30}+\frac{1}{15}\left(\frac{ab+bc+ca}{a^2+b^2+c^2}\right)\left(2\right)\).Từ (1);(2) có:

\(P=\frac{1}{30}-\frac{3}{2}+\frac{1}{5}\left(\frac{ab+bc+ca}{a^2+b^2+c^2}\right)+\frac{9}{4}\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)-\frac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}\)

\(=\frac{1}{15}\left(\frac{a^2+b^2+c^2}{ab+bc+ac}+\frac{ab+bc+ca}{a^2+b^2+c^2}-22\right)\ge-\frac{4}{3}\)

đề thi hsg toán lớp 9 tỉnh thanh hóa năm 2016-2017 mà

\(=\left(2x^2+x+1\right)^2\)

mịa c đâu ra vậy

Ta có :

\(a-\sqrt{a}+\frac{1}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2\ge0\forall a\ge0\Rightarrow a+\frac{1}{4}\ge\sqrt{a}\)

\(b-\sqrt{b}+\frac{1}{4}=\left(\sqrt{b}-\frac{1}{2}\right)^2\ge0\forall b\ge0\Rightarrow b+\frac{1}{4}\ge\sqrt{b}\)

\(\Rightarrow a+\frac{1}{4}+b+\frac{1}{4}\ge\sqrt{a}+\sqrt{b}\)

\(\Rightarrow a+b+\frac{1}{2}\ge\sqrt{a}+\sqrt{b}\)(đpcm)

a)

\(3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}^2\right)+2\times\sqrt{2}\times1=\left(\sqrt{2}+1\right)^2\)

mấy câu còn lại tương tự