cái này là bđt bunhia thì fai bn mở sách ra tham khảo đi

a/ \(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\ge\left(x+y\right)\left(2xy-xy\right)=xy\left(x+y\right)\)

Dấu "=" xảy ra khi \(x=y\)

b/Đặt biểu thức vế trái là Q

\(\frac{1}{a+b+1+3}\le\frac{1}{4}\left(\frac{1}{a+b+1}+\frac{1}{3}\right)=\frac{1}{4}\left(\frac{1}{a+b+1}\right)+\frac{1}{12}\)

Thiết lập tương tự và cộng lại:

\(Q\le\frac{1}{4}\left(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\right)+\frac{1}{4}\)

Xét \(P=\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\)

Đặt \(\left(a;b;c\right)=\left(x^3;y^3;z^3\right)\Rightarrow xyz=1\)

\(\Rightarrow P=\frac{1}{x^3+y^3+1}+\frac{1}{y^3+z^3+1}+\frac{1}{z^3+x^3+1}\le\frac{1}{xy\left(x+y\right)+1}+\frac{1}{yz\left(y+z\right)+1}+\frac{1}{zx\left(z+x\right)+1}\)

\(P\le\frac{xyz}{xy\left(x+y\right)+xyz}+\frac{xyz}{yz\left(y+z\right)+xyz}+\frac{xyz}{zx\left(z+x\right)+xyz}\)

\(P\le\frac{z}{x+y+z}+\frac{x}{x+y+z}+\frac{y}{x+y+z}=1\)

\(\Rightarrow Q\le\frac{1}{4}.1+\frac{1}{4}=\frac{1}{2}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

Câu 2: \(\left(\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\right)^2=\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{xz}{y}\right)^2+2\left(x^2+y^2+z^2\right)\)

\(=\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{xz}{y}\right)^2+6\)

Áp dụng bất đẳng thức AM - GM ta có :

\(\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{xz}{y}\right)^2\ge3\sqrt[3]{\left(\frac{xy}{z}\right)^2\left(\frac{yz}{x}\right)^2\left(\frac{xy}{y}\right)^2}=3\sqrt[3]{\frac{\left(xyz\right)^4}{\left(xyz\right)^2}}=3\)\(\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\ge\sqrt{3+6}=3\left(dpcm\right)\)

tại sao lại suy ra đc \(3\sqrt[3]{\frac{\left(xyz\right)^4}{\left(xyz\right)^{^2}}}=3\) vậy cậu?

Bài 1:

a: Sửa đề: \(\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}=0\)

Ta có: \(\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}\)

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

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

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

\(=\frac{-1}{2ab}+\frac{-1}{2bc}+\frac{-1}{2ac}=\frac{-c-a-b}{2abc}=0\)

2:

Sửa đề: \(M=\frac{x^2}{x^4+x^2+1}\)

TH1: x=0

=>\(a=\frac{x}{x^2+x+1}=0\)

\(M=\frac{x^2}{x^4+x^2+1}=0\)

TH2: x<>0

\(\frac{x}{x^2+x+1}=a\)

=>\(\frac{x^2+x+1}{x}=\frac{1}{a}\)

=>\(\frac{x^2-x+1+2x}{x}=\frac{1}{a}\)

=>\(\frac{x^2-x+1}{x}+2=\frac{1}{a}\)

=>\(\frac{x^2-x+1}{x}=\frac{1}{a}-2=\frac{1-2a}{a}\)

=>\(\frac{x}{x^2-x+1}=\frac{a}{1-2a}\)

\(M=\frac{x^2}{x^4+x^2+1}\)

\(=\frac{x^2}{x^4+2x^2+1-x^2}\)

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

\(=\frac{x}{x^2-x+1}\cdot\frac{x}{x^2+x+1}=a\cdot\frac{a}{1-2a}=\frac{a^2}{1-2a}\)

post ít một thôi