áp dụng bất đẳng thức: 1+b²>=2b. tương tự.....

ad bđt cauchy: a/b+b/c+c/a>=3∛a/b.b/c.c/a=3

P>=\(\dfrac{2ab}{bc}\)+\(\dfrac{2bc}{ca}\)+\(\dfrac{2ca}{ab}\) =2(\(\dfrac{a}{b}\)+\(\dfrac{b}{c}\)+ \(\dfrac{c}{a}\))>=2.3=6

Pmin khi a=b=c=1

Áp dụng bđt : \(1+b^2>=2b\)

bđt cauchy : \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}>3\sqrt[3]{}\) a\b . b\c . c\a = 3

\(P=2+\dfrac{2}{b}+a+\dfrac{a}{b}+2+\dfrac{2}{a}+b+\dfrac{b}{a}=\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(a+\dfrac{1}{2a}\right)+\left(b+\dfrac{1}{2b}\right)+\left(\dfrac{3}{2a}+\dfrac{3}{2b}\right)+4\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}+2\sqrt{a.\dfrac{1}{2a}}+2\sqrt{b.\dfrac{1}{2b}}+2\sqrt{\dfrac{3}{2a}.\dfrac{3}{2b}}+4=6+2\sqrt{2}+\dfrac{3}{\sqrt{ab}}\)

Ta lại có: \(a^2+b^2\ge2\sqrt{a^2.b^2}=2ab\left(BĐT.Cauchy\right)\Rightarrow2\left(a^2+b^2\right)\ge4ab\Rightarrow\sqrt{ab}\le\dfrac{\sqrt{2\left(a^2+b^2\right)}}{2}=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow P\ge6+2\sqrt{2}+\dfrac{3}{\sqrt{ab}}\ge6+2\sqrt{2}+\dfrac{3}{\dfrac{\sqrt{2}}{2}}=6+5\sqrt{2}\)

\(minP=6+5\sqrt{2}\Leftrightarrow a=b=\dfrac{\sqrt{2}}{2}\)

Tử, mẫu không đồng bậc

Đề sai hoặc thiếu điều kiện

tử cộng thêm c^2 bớts c^2

tách tử theo mẫu

cô si mẫu

Đặt \(\dfrac{b}{a}=x;\dfrac{c}{b}=y\).

Ta có: \(P=\dfrac{1}{\left(\dfrac{a+b}{a}\right)^2}+\dfrac{1}{\left(\dfrac{b+c}{b}\right)^2}+\dfrac{b}{a}.\dfrac{c}{b}.\dfrac{1}{4}\)

\(P=\dfrac{1}{\left(x+1\right)^2}+\dfrac{1}{\left(y+1\right)^2}+\dfrac{xy}{4}\).

Ta có bđt quen thuộc: \(\dfrac{1}{\left(x+1\right)^2}+\dfrac{1}{\left(y+1\right)^2}\ge\dfrac{1}{xy+1}\) (bạn xem cm ở đây).

Do đó \(P\ge\dfrac{1}{xy+1}+\dfrac{xy+1}{4}-\dfrac{1}{4}\ge1-\dfrac{1}{4}=\dfrac{3}{4}\).

Đẳng thức xảy ra khi x = y = 1 tức a = b = c. 

Vậy...

\(A=a^3b^3+\dfrac{1}{a^3b^3}+2=a^3b^3+\dfrac{1}{2^{12}.a^3b^3}+\dfrac{2^{12}-1}{2^{12}a^3b^3}+2\)

\(A\ge2\sqrt{\dfrac{a^3b^3}{2^{12}.a^3b^3}}+\dfrac{2^{12}-1}{2^{12}.\left(\dfrac{a+b}{2}\right)^6}+2=\dfrac{2}{2^6}+\dfrac{2^{12}-1}{2^6}+2=\dfrac{2^{12}+1}{2^6}+2\) (casio)

Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)

Biểu thức này chỉ có max, không có min

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

Vì \(a^2+b^2=1\) và \(a,b>0\Leftrightarrow0< a< 1;0< b< 1\Leftrightarrow1+a-b>0;1-b+a>0\)

\(\Leftrightarrow A\ge2\sqrt{\dfrac{\left(1-a+b\right)\left(1-b+a\right)}{ab}}=2\sqrt{\dfrac{1-a^2-b^2+2ab}{ab}}=2\sqrt{2}\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\dfrac{1-a+b}{b}=\dfrac{1-b+a}{a}\end{matrix}\right.\Leftrightarrow a=b=\dfrac{1}{\sqrt{2}}\)