BĐT cần  chứng minh tương đương với :

\(\left(a^2b+b^2c+c^2a\right)\left(2+\frac{1}{a^2b^2c^2}\right)\ge9\)

\(\Leftrightarrow2\left(a^2b+b^2c+c^2a\right)+\frac{1}{ab^2}+\frac{1}{bc^2}+\frac{1}{ca^2}\ge9\)

Áp dụng BĐT Cô-si cho 3 số dương ,ta có :

\(a^2b+a^2b+\frac{1}{ab^2}\ge3\sqrt[3]{a^2b.a^2b.\frac{1}{ab^2}}=3a\)

tương tự :  \(b^2c+bc^2+\frac{1}{bc^2}\ge3b\), \(\left(c^2a+ca^2+\frac{1}{ca^2}\right)\ge3c\)

Cộng 3 BĐT trên theo vế, ta được :

\(2\left(a^2b+b^2c+c^2a\right)+\frac{1}{ab^2}+\frac{1}{bc^2}+\frac{1}{ca^2}\ge3\left(a+b+c\right)=9\)

Dấu "=" xảy ra khi a = b = c = 1

Ta có:

\(VP=4\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\left(\frac{1-a+1-c}{2}\right)^2\cdot4\left(1-b\right)=\left(2-a-c\right)\left(2-a-c\right)\left(1-b\right)\le\left(a+2b+c\right)\left(\frac{2-a-c+1-b}{2}\right)^2=\left(\frac{2+1-1}{2}\right)^2\left(a+2b+c\right)=VT\)

Đấu "=" khi \(\left\{{}\begin{matrix}a+b+c=1\\1-a=1-c\\2-a-c=1-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=c\\2a+b=1\\1-2a=-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1-2a=b\\1-2a=-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=0\\a=c=\frac{1}{2}\end{matrix}\right.\)

Sửa \(\le\) thành \(\ge\) nha bạn

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)

Ta có \(\dfrac{a^2}{a+bc}=\dfrac{a^3}{a^2+abc}=\dfrac{a^3}{a^2+ab+bc+ca}=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}\)

Tương tự: \(\left\{{}\begin{matrix}\dfrac{b^2}{b+ca}=\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}\\\dfrac{c^2}{c+ba}=\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\end{matrix}\right.\)

Áp dụng BĐT cosi:

\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3}{4}a\)

\(\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{b^3}{64}}=\dfrac{3}{4}b\)

\(\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{c^3}{64}}=\dfrac{3}{4}c\)

Cộng VTV:

\(\Leftrightarrow VT+\dfrac{a+b}{8}+\dfrac{a+c}{8}+\dfrac{b+c}{8}\ge\dfrac{3}{4}\left(a+b+c\right)\\ \Leftrightarrow VT\ge\dfrac{3\left(a+b+c\right)}{4}-\dfrac{2\left(a+b+c\right)}{8}\\ \Leftrightarrow VT\ge\dfrac{a+b+c}{4}\)

Dấu \("="\Leftrightarrow a=b=c=3\)

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)

BĐT trở thành: \(\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt{x^2+z^2+2y^2}}\le\dfrac{1}{2}\)

Ta có:

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

\(\Rightarrow\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)

Tương tự: \(\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}\le\dfrac{1}{2}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)

\(\dfrac{zx}{\sqrt{z^2+x^2+2y^2}}\le\dfrac{1}{2}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)\)

Cộng vế với vế:

\(VT\le\dfrac{1}{2}\left(\dfrac{zx+yz}{x+y}+\dfrac{xy+zx}{y+z}+\dfrac{yz+xy}{z+x}\right)=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)

ôn zăn ik bà, tón gánh hong nổi đâu =))

bài 1b

+)Nếu n chẵn ,ta có \(n^4⋮2,4^n⋮2\Rightarrow n^4+4^n⋮2\)

mà \(n^4+4^n>2\)Do đó \(n^4+4^n\)là hợp số

+)nếu n lẻ đặt \(n=2k+1\left(k\in N\right)\)

Ta có \(n^4+4^n=n^4+4^{2k}.4=\left(n^2+2.4k\right)^2-2n^2.2.4^k\)

\(=\left(n^2+2^{2k+1}\right)^2-\left(2.n.2^k\right)^2\)

\(=\left(n^2+2^{2k+1}+2n.2^k\right)\left(n^2+2^{2k+1}-2n.2^k\right)\)

\(=\left(\left(n+2^k\right)^2+2^{2k}\right)\left(\left(n-2^k\right)^2+2^{2k}\right)\)

là hợp số,vì mỗi thừa số đều lớn hơn hoặc bằng 2

(nhớ k nhé)

Bài 2a)

Nhân 2 vế với 2 ta có

\(a^4+b^4\ge2ab\left(a^2+b^2\right)-2a^2b^2\)

\(\Leftrightarrow\left(a^2+b^2\right)^2\ge2ab\left(a^2+b^2\right)\)

\(\Leftrightarrow a^2+b^2\ge2ab\Leftrightarrow\left(a-b\right)^2\ge0\)(đúng)

Dẫu = xảy ra khi \(a=b\)

dat a+b=x b+c=y c+a=z \(\Rightarrow\) dt tro thanh \(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=2\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{1}{y+1}+1-\frac{1}{z+1}=\frac{y}{y+1}+\frac{z}{z+1}\) \(\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\) (bdt amgm)

tuong tu \(\frac{1}{y+1}\ge2\sqrt{\frac{xz}{\left(x+1\right)\left(z+1\right)}}\) \(\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\)

\(\frac{\Rightarrow1}{x+1}.\frac{1}{y+1}.\frac{1}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}.2\sqrt{\frac{xz}{\left(z+1\right)\left(x+1\right)}}.2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\)

                =\(8.\frac{xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(\Rightarrow xyz\le\frac{1}{8}\)dau = xay ra khi x=y=z=1/2 hay a=b=c=1/4