b) \(\dfrac{1}{3a+2b+c}\le\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{1}{36}\left(\dfrac{3}{a}+\dfrac{2}{b}+\dfrac{1}{c}\right)\)

Tương tự cho 2 cái kia rồi cộng lại

\(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}.16=\dfrac{8}{3}\)

Đẳng thức xảy ra \(\Leftrightarrow\) ... \(\Leftrightarrow a=b=c=\dfrac{3}{16}\)

p,q là gì??

ĐK: \(x,y,z\ne0\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=0\Leftrightarrow xy+xz+yz=0\)

\(\Rightarrow\left\{{}\begin{matrix}xy=-xz-yz\\xz--xy-yz\\yz=-xy-xz\end{matrix}\right.\)

Ta có:

\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-xz=x\left(x-y\right)-z\left(x-y\right)\)

\(=\left(x-y\right)\left(x-z\right)\Rightarrow\dfrac{1}{x^2+2yz}=\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)

Tương tự: \(\dfrac{1}{y^2+2xz}=\dfrac{1}{\left(y-x\right)\left(y-z\right)}=\dfrac{-1}{\left(x-y\right)\left(y-z\right)}\)

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

Cộng vế với vế ta được:

\(\dfrac{1}{x^2+2yz}+\dfrac{1}{y^2+2xz}+\dfrac{1}{z^2+2xy}=\dfrac{1}{\left(x-y\right)\left(x-z\right)}+\dfrac{-1}{\left(x-y\right)\left(y-z\right)}+\dfrac{1}{\left(x-z\right)\left(y-z\right)}\)

\(=\dfrac{y-z-\left(x-z\right)+x-y}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{y-z-x+z+x-y}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=0\) (đpcm)

\(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xy}+\dfrac{2}{yz}+\dfrac{2}{xz}-\dfrac{2}{xy}-\dfrac{2}{yz}-\dfrac{2}{xz}}=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)}=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2.\dfrac{x+y+z}{xyz}}=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\text{|}\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\text{|}\)

\(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)-2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)}\\ =\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\left(\dfrac{z}{xyz}+\dfrac{y}{xyz}+\dfrac{x}{xyz}\right)}\\ =\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+x+y+z)\geq (1+1+1+1)^2\)

\(\Rightarrow \frac{2}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{16}{2x+y+z}\)

Hoàn toàn tương tự:

\(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\geq \frac{16}{x+2y+z}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\geq \frac{16}{x+y+2z}\)

Cộng theo vế các BĐT vừa thu được:

\(\Rightarrow 4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq 16\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)

\(\Rightarrow 16\geq 16\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)

\(\Rightarrow \frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\leq 1\)

Ta có đpcm.

Ta có :

\(\dfrac{1}{2x+y+z}=\dfrac{16}{16\left(x+x+y+z\right)}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(\dfrac{1}{x+2y+z}=\dfrac{16}{16\left(x+y+y+z\right)}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(\dfrac{1}{x+y+2z}=\dfrac{16}{16\left(x+y+z+z\right)}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}\right)\)

Cộng từng vế của BĐT ta được :

\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)

Vậy BĐT đã được chứng minh !

2a)

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2a+b+c}=\dfrac{1}{a+b+a+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\\dfrac{1}{a+2b+c}=\dfrac{1}{a+b+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\\\dfrac{1}{a+b+2c}=\dfrac{1}{a+c+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{1}{4}\left(\dfrac{1}{b+c}+\dfrac{1}{a+b}\right)+\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)

\(\Rightarrow VT\le\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}\)

\(\Rightarrow VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)

Chứng minh rằng \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

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

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\\\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\)

\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )

Vì \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Mà \(VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)

\(\Rightarrow\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)( đpcm )

Dấu " = " xảy ra khi \(a=b=c\)

2b)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}1+a^2\ge2\sqrt{a^2}=2a\\1+b^2\ge2\sqrt{b^2}=2b\\1+c^2\ge2\sqrt{c^2}=2c\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{1+a^2}\le\dfrac{a}{2a}=\dfrac{1}{2}\\\dfrac{b}{1+b^2}\le\dfrac{b}{2b}=\dfrac{1}{2}\\\dfrac{c}{1+c^2}\le\dfrac{c}{2c}=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\le\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c=1\)

Bài 1)

Nháp : nhìn nhanh ta thấy nên áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Giải

Vì x,y > 0 =) 2x + y > 0 , x + 2y > 0

Áp dụng BĐT cauchy dạng phân thức cho hai bộ số không âm \(\dfrac{1}{2x+y}\)và\(\dfrac{1}{x+2y}\)

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

\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3\left(x+y\right)}=4\)

Dấu '' = "xảy ra khi và chỉ khi x + 2y = y + 2x (=) x=y

ta có:

\(1+a^2\ge2\sqrt{a^2.1}=2a\Rightarrow\dfrac{a}{1+a^2}\le\dfrac{a}{2a}=\dfrac{1}{2}\)

tương tự, ta có: \(\dfrac{b}{b^2+1}\le\dfrac{1}{2}\); \(\dfrac{c}{c^2+1}\le\dfrac{1}{2}\)

công từ vế với nhau, ta có điều cần phải chứng minh

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

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

mà a+b+c=0

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{0}{abc}\right)=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)

3/ Áp dụng bất đẳng thức AM-GM, ta có :

\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)

\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)

\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)

Cộng 3 vế của BĐT trên ta có :

\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)

\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)

Bài 1:

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)

Tiếp tục áp dụng BĐT AM-GM:

\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)

Do đó:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$

Bài 2:

Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)

\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)

\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)

Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$