ÁP dụng AM-GM:

\(\sum\dfrac{a^2}{\sqrt{1-a^2}}=\sum\dfrac{a^3}{\sqrt{\left(1-a^2\right).a^2}}\ge\sum\dfrac{a^3}{\dfrac{1}{2}\left(1-a^2+a^2\right)}=2\sum a^3=2\left(đpcm\right)\)

Dấu = không xảy ra

Text

\(VT=\dfrac{3}{xy+yz+xz}+\dfrac{2}{x^2+y^2+z^2}\)

\(=\dfrac{8}{4\left(xy+yz+xz\right)}+\dfrac{4}{4\left(xy+yz+xz\right)}+\dfrac{4}{2\left(x^2+y^2+z^2\right)}\)

\(\ge\dfrac{8}{4\cdot\dfrac{\left(x+y+z\right)^2}{3}}+\dfrac{\left(2+2\right)^2}{2\left(x+y+z\right)^2}\)

\(=\dfrac{8}{4\cdot\dfrac{1^2}{3}}+\dfrac{\left(2+2\right)^2}{2\cdot1^2}=14\)

\("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)

\(x^2-xy+y^2+1>0\)

\(\Leftrightarrow x^2-xy+\frac{1}{4}y^2+\frac{3}{4}y^2+1>0\)

\(\Leftrightarrow\left(x^2-xy+\frac{1}{4}y^2\right)+\frac{3}{4}y^2+1>0\)

\(\Leftrightarrow\left[x^2-2\cdot x\cdot\frac{1}{2}y+\left(\frac{1}{2}y\right)^2\right]+\frac{3}{4}y^2+1>0\)

\(\Leftrightarrow\left(x-\frac{1}{2}y\right)^2+\frac{3}{4}y^2+1>0\)( đúng với ∀ x, y ∈ R )

=> đpcm 

Ta có : \(x^2+2y^2+2xy+y+1\)

\(=\left(x^2+2xy+y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\)

\(=\left(x+y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x,y\)

lớp 8 thì còn lằng nhằng lớp 10 quá đơn giản

\(x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{1}{3}\)

Dùng phép biến đổi tương đương, ta có:

x^2 + y^2 + x^2 >= 1/3
<=> x^2 + y^2 + x^2 >= (x + y + z)/3 ( vì x + y + z = 1)
<=> x^2 + y^2 + x^2 - (x + y + z)/3 >= 0
<=> 3x^2 + 3y^2 + 3z^2 - x - y - z >= 0
<=> x(3x - 1) + y(3y - 1) + z(3z - 1) >= 0
<=> x(3x - x - y - z) + y(3y - x - y - z) + z(3z - x - y - z) >= 0
<=> x(2x - y - z) + y(2y - x -z) + z(2z - x - y) >= 0
<=> 2x^2 - xy - xz + 2y^2 - xy - yz + 2z^2 - xz - yz >= 0
<=> (x^2 - 2xy - y^2) + (y^2 - 2yz - z^2) + (x^2 - 2xz - z^2) >= 0
<=> (x - y)^2 + (y - z)^2 - (x - z)^2 >= 0 (đúng)
=> x^2 + y^2 + x^2 >= 1/3

Dấu = xảy ra <=> x = y = z =1/3

BĐT bạn ghi ngược rồi, BĐT đúng phải là:

\(\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{8}{\left(x+y\right)^2}\)

Chứng minh:

Ta có: \(\left(\frac{1}{x}\right)^2+\left(\frac{1}{y}\right)^2\ge\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)^2\ge\frac{1}{2}\left(\frac{4}{x+y}\right)^2=\frac{8}{\left(x+y\right)^2}\)

Dấu "=" xảy ra khi \(x=y\)

áp dụng bđt dang Engel

P=1/[x(x+y) ]+1/[y(x+y) ]

=1/(x+y). (1/x+1/y)

=1/(x+y). [(x+y) /xy]=1/(xy)

x+y≤1,x, y>0=>x.y≤1/4

p≥1/(1/4)=4

đẳng thức khi x=y=1/2