em lp 6  a ơi

BĐT bên trái rất đơn giản, chỉ cần áp dụng:

\(x^3+x^3+y^3\ge3x^2y\) ; tương tự và cộng lại và được

Ta chứng minh BĐT bên phải:

\(\Leftrightarrow x^4+y^4+z^4+2\ge2\left(x^3+y^3+z^3\right)=\left(x+y+z\right)\left(x^3+y^3+z^3\right)\)

\(\Leftrightarrow2\ge x^3\left(y+z\right)+y^3\left(z+x\right)+z^3\left(x+y\right)\)

\(\Leftrightarrow\dfrac{1}{8}\left(x+y+z\right)^4\ge x^3\left(y+z\right)+y^3\left(z+x\right)+z^3\left(x+y\right)\)

Thật vậy, ta có:

\(\dfrac{1}{8}\left(x+y+z\right)^4=\dfrac{1}{8}\left[x^2+y^2+z^2+2\left(xy+yz+zx\right)\right]^2\)

\(\ge\dfrac{1}{8}.4\left(x^2+y^2+z^2\right).2\left(xy+yz+zx\right)=\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)\)

\(=x^3\left(y+z\right)+y^3\left(z+x\right)+z^3\left(x+y\right)+xyz\left(x+y+z\right)\)

\(\ge x^3\left(y+z\right)+y^3\left(z+x\right)+z^3\left(x+y\right)\) (đpcm)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(0;1;1\right)\) và hoán vị

Ta đặt: \(\left\{{}\begin{matrix}\dfrac{1}{x^2}=a\\\dfrac{1}{y^2}=b\\\dfrac{1}{z^2}=c\end{matrix}\right.\)\(\Rightarrow\sqrt{abc}=abc=1\)

Ta có: \(\dfrac{1}{\sqrt{a}+\sqrt{ab}+1}+\dfrac{1}{\sqrt{b}+\sqrt{bc}+1}+\dfrac{1}{\sqrt{c}+\sqrt{ca}+1}\)

\(=\dfrac{1}{\sqrt{a}+\sqrt{ab}+1}+\dfrac{1}{\sqrt{b}+\dfrac{1}{\sqrt{a}}+1}+\dfrac{1}{\dfrac{1}{\sqrt{ab}}+\sqrt{ca}+1}\)

\(=\dfrac{1}{\sqrt{a}+\sqrt{ab}+1}+\dfrac{\sqrt{a}}{\sqrt{ba}+1+\sqrt{a}}+\dfrac{1}{1+\sqrt{ab}+\sqrt{a}}=1\)

Quay lại bài toán, sau khi đặt bài toán trở thành:

\(P=\dfrac{1}{2b+a+3}+\dfrac{1}{2c+b+3}+\dfrac{1}{2a+c+3}\)

\(=\dfrac{1}{\left(a+b\right)+\left(b+1\right)+2}+\dfrac{1}{\left(b+c\right)+\left(c+1\right)+2}+\dfrac{1}{\left(c+a\right)+\left(a+1\right)+2}\)

\(\le\dfrac{1}{2}\left(\dfrac{1}{\sqrt{a}+\sqrt{ab}+1}+\dfrac{1}{\sqrt{b}+\sqrt{bc}+1}+\dfrac{1}{\sqrt{c}+\sqrt{ca}+1}\right)=\dfrac{1}{2}\)

Cái đó t cố tình bỏ đấy. B phải tự làm chứ chẳng lẽ t làm hết??

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

\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}>=2\cdot\sqrt{\dfrac{y^2}{x^2}\cdot\dfrac{x^2}{y^2}}=2\)

=>\(VT>=5+\left(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}\right)+\left(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}\right)+\dfrac{15}{16}z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)

\(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}>=2\cdot\sqrt{\dfrac{x^2}{z^2}\cdot\dfrac{z^2}{16x^2}}=\dfrac{1}{2}\)

\(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}>=\dfrac{1}{2}\)

và \(\dfrac{1}{x^2}+\dfrac{1}{y^2}>=\dfrac{2}{xy}>=\dfrac{2}{\left(\dfrac{x+y}{2}\right)^2}=\dfrac{8}{\left(x+y\right)^2}\)

=>\(\dfrac{15}{16}z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)>=\dfrac{15}{16}z^2\cdot\dfrac{8}{\left(x+y\right)^2}=\dfrac{15}{2}\left(\dfrac{z}{x+y}\right)^2=\dfrac{15}{2}\)

=>VT>=5+1/2+1/2+15/2=27/2

\(\frac{1}{x^4}+\frac{1}{y^4}=\frac{x^2}{x^6}+\frac{1}{y^4}\ge\frac{\left(x+1\right)^2}{x^6+y^4}\ge\frac{4x}{x^6+y^4}\)

tương tự

\(\frac{1}{y^4}+\frac{1}{z^4}\ge\frac{4y}{y^6+z^4}\);

\(\frac{1}{z^4}+\frac{1}{x^4}\ge\frac{4z}{z^6+x^4}\);

cộng vế với vế => đpcm

Dấu "=" xảy ra <=> x=y=z=1

Cô si

Cách khác:

\(x^6+y^4\ge2\sqrt{x^6y^4}=2x^3y^2\)

\(\Rightarrow\frac{2}{x^6+y^4}\le\frac{1}{x^2y^2}\)

CMTT , ta có VT \(\le\frac{1}{x^2y^2}+\frac{1}{y^2z^2}+\frac{1}{z^2x^2}\)

Bổ đề: \(a^2+b^2+c^2\ge ab+bc+ca\) ( luôn đúng)

\(\Rightarrow\frac{1}{x^2y^2}+\frac{1}{y^2z^2}+\frac{1}{z^2x^2}\le\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\)

ĐPCM

Dấu " =" xảy ra khi x=y=z=1