Bài 1:

\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\) với a,b,c > 0

Áp dụng BĐT Chauchy cho 2 số không âm, ta có:

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

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

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

Cộng vế theo vế ta được:

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

\(\Leftrightarrow\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\)

Đề sai, ví dụ với \(\left(a;b;c\right)=\left(3;3;2\right)\) thì vế trái xấp xỉ \(2.78< \dfrac{11930}{2821}\)

Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)

\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)

CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)

Cộng vế theo vế:

\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

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

= \(1+\dfrac{a}{b}+\dfrac{b}{a}+1\)

=\(2+\dfrac{a}{b}+\dfrac{b}{a}\)

áp dụng BĐT cô si cho 2 số ta có

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)

=> \(2+\dfrac{a}{b}+\dfrac{b}{a}\ge4\)

<=> \(\left(a+b\right)\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge4\)(đpcm)

bn vô câu hỏi tương tự có hết nhé

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(\text{VT}=\frac{\left ( \frac{a}{bc} \right )^2}{\frac{1}{c}}+\frac{\left ( \frac{b}{ca} \right )^2}{\frac{1}{a}}+\frac{\left ( \frac{c}{ab} \right )^2}{\frac{1}{b}}\geq \frac{\left ( \frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\)

\(\Leftrightarrow \text{VT}\geq \frac{\left ( \frac{a^2+b^2+c^2}{abc} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\)

Theo hệ quả của BĐT AM-GM thì:

\(a^2+b^2+c^2\geq ab+bc+ac\)

\(\Rightarrow \text{VT}\geq \frac{\left ( \frac{ab+bc+ac}{abc} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Ta có đpcm.

Dấu bằng xảy ra khi \(a=b=c\)

Áp dụngk BĐt cô-si, ta có 

\(\frac{a^2}{b^2c}+\frac{b^2}{c^2a}+\frac{1}{a}\ge3.\frac{1}{c}\)

Tương tự , rồi cộng vào, ta có 

\(2A+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\Rightarrow A\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\left(ĐPCM\right)\)

^_^ 

\(vì:a,b,c>0\Rightarrow\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}>0\)

\(Cosi:\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{2}{\sqrt{ab}}\ge\dfrac{2}{\dfrac{a+b}{2}}=\dfrac{4}{a+b}\)

\(\dfrac{4}{2a+b+c}\le\dfrac{1}{4}\left(\dfrac{4}{a+b}+\dfrac{4}{a+c}\right)\le\dfrac{1}{16}\left(\dfrac{8}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=\dfrac{1}{2a}+\dfrac{1}{4b}+\dfrac{1}{4c}.tươngtự:\dfrac{4}{a+b+2c}\le\dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{2c};\dfrac{4}{a+2b+c}\le\dfrac{1}{4a}+\dfrac{1}{2b}+\dfrac{1}{2c}.\text{cộng vế theo vế ta được:}\dfrac{4}{a+2b+c}+\dfrac{4}{2a+b+c}+\dfrac{4}{a+b+2c}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(\text{đpcm}\right)\)

Áp dụng BĐT \(\dfrac{1}{x+y+z+t}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)\) với các số dương

Ta có: \(\dfrac{4}{a+a+b+c}\le\dfrac{4}{16}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

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

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

Cộng vế với vế:

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

Dấu "=" xảy ra khi \(a=b=c\)

* Ta cm bđt : \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\forall ab\)

+ \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

\(\Leftrightarrow\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\)

\(\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow\left(a+b\right)^2-4ab\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\)

Vì bđt thức cuối luôn đúng mà các phép biến đổi trên là tương đương nên ta có đpcm

Dấu "=" \(\Leftrightarrow x=y\)

+ Áp dụng bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Dấu "=" \(\Leftrightarrow x=y\) ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) Dấu "=" xảy ra \(\Leftrightarrow a=b\)

\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\) Dấu "=" xảy ra \(\Leftrightarrow b=c\)

\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{a+c}\) Dấu "=" xảy ra \(\Leftrightarrow a=c\)

Do đó : \(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge4\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge2\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)

+ Áp dụng bđt trên một lần nữa ta có :

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}\ge\dfrac{4}{a+2b+c}\) Dấu "=" xảy ra \(\Leftrightarrow a=c\)

\(\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{1}{a+b+2c}\) Dấu "=" xảy ra \(\Leftrightarrow a=b\)

\(\dfrac{1}{a+b}+\dfrac{1}{c+a}\ge\dfrac{4}{2a+b+c}\) Dấu "=" xảy ra \(\Leftrightarrow b=c\)

Do đó : \(2\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{4}{2a+b+c}\)

\(+\dfrac{4}{a+2b+c}+\dfrac{4}{a+b+2c}\)

=> đpcm

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)