K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

8 tháng 4 2022

ĐK : a;b;c > 0 

Ta có : \(ab+bc+ac=1\) \(\Leftrightarrow c\left(a+b\right)=1-ab\Leftrightarrow c=\dfrac{1-ab}{a+b}\)

Khi đó :  \(c^2+1=\left(\dfrac{1-ab}{a+b}\right)^2+1\)  \(=\dfrac{\left(ab\right)^2+1+a^2+b^2}{\left(a+b\right)^2}=\dfrac{\left(a^2+1\right)\left(b^2+1\right)}{\left(a+b\right)^2}\)

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

Ta có : \(\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}=\dfrac{ab^2+a^2b+a+b}{\left(a^2+1\right)\left(b^2+1\right)}=\dfrac{\left(ab+1\right)\left(a+b\right)}{\left(a^2+1\right)\left(b^2+1\right)}\)

Suy ra : \(A=\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}-\dfrac{1}{c^2+1}=\dfrac{\left(a+b\right)\left(ab+1-a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)}=\dfrac{\left(a+b\right)\left(1-a\right)\left(1-b\right)}{\left(a^2+1\right)\left(b^2+1\right)}\)

AD BĐT Cauchy ta được :  \(\left(a+b\right)\left[\left(1-a\right)\left(1-b\right)\right]\le\dfrac{\left[a+b+\left(1-a\right)\left(1-b\right)\right]^2}{4}=\dfrac{\left(1+ab\right)^2}{4}\)

\(\left(a^2+1\right)\left(b^2+1\right)\ge\left(ab+1\right)^2\)  ( theo BCS )

Suy ra : \(A\le\dfrac{1}{4}\)

26 tháng 1 2022

\(M=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}\)

\(M\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}+\dfrac{7}{ab+bc+ca}=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)

\(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\)

\(\Rightarrow M\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}=9+\dfrac{7.3}{\left(a+b+c\right)^2}=9+21=30\)

\(Min_M=30\Leftrightarrow a=b=c=\dfrac{1}{3}\)

26 tháng 1 2022

Áp dụng BĐT Svacxo

\(m\text{≥}\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}\)

\(=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}\)

≥ \(\dfrac{9}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}\)\(+\dfrac{7}{ab+bc+ca}\)

\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)

CM BĐT: \(a^2+b^2+c^2\text{≥}ab+bc+ca\)

⇔ \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\text{≥}0\) (luôn đúng)

⇒ \(\left(a+b+c\right)^2\text{≥}3\left(ab+bc+ca\right)\)

⇒ \(\dfrac{\left(a+b+c\right)^2}{3}\text{≥}ab+bc+ca\)

⇒ \(m\text{≥}\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{\dfrac{\left(a+b+c\right)^2}{3}}=9+21=30\) 

(vì a+b+c=1)

Vậy...

AH
Akai Haruma
Giáo viên
5 tháng 2

Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(P=2(\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2})+\frac{1}{2(ab+bc+ac)}\\ \geq 2.\frac{9}{2(ab+bc+ac)+a^2+b^2+c^2}+\frac{1}{2(ab+bc+ac)}\\ =\frac{18}{(a+b+c)^2}+\frac{1}{2(ab+bc+ac)}\\ =18+\frac{1}{2(ab+bc+ac)}\)

Áp dụng BĐT AM-GM:

$2(ab+bc+ac)\leq 2.\frac{(a+b+c)^2}{3}=\frac{2}{3}$

$\Rightarrow \frac{1}{2(ab+bc+ac)}\geq \frac{3}{2}$

$\Rightarrow P\geq 18+\frac{3}{2}=\frac{39}{2}$
Vậậy $P_{\min}=\frac{39}{2}$ khi $a=b=c=\frac{1}{3}$

27 tháng 4 2023

Với \(ab+bc+ca=1\) và a,b,c>0 ta có:

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

\(\dfrac{\sqrt{a^2+1}.\sqrt{b^2+1}}{\sqrt{c^2+1}}=a+b\)

Tương tự: \(\dfrac{\sqrt{b^2+1}.\sqrt{c^2+1}}{\sqrt{a^2+1}}=b+c\) ; \(\dfrac{\sqrt{c^2+1}.\sqrt{a^2+1}}{\sqrt{b^2+1}}=c+a\)

\(\Rightarrow P=2\left(a+b+c\right)\)

\(\Rightarrow P^2=4\left(a+b+c\right)^2\ge4.3\left(ab+bc+ca\right)=4.3.1=12\)

\(\Rightarrow P\ge2\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{\sqrt{3}}{3}\)

Vậy \(MinP=2\sqrt{3}\)

18 tháng 4 2022

a,b,c >=0 à bạn 

8 tháng 1 2022

\(\sqrt{\dfrac{ab}{c+ab}}=\sqrt{\dfrac{ab}{1-a-b-ab}}=\sqrt{\dfrac{ab}{\left(1-b\right)\left(1-a\right)}}\le\dfrac{\dfrac{a}{1-b}+\dfrac{b}{1-a}}{2}\left(1\right)\) \(tương-tự\Rightarrow\sqrt{\dfrac{bc}{a+bc}}\le\dfrac{\dfrac{b}{1-c}+\dfrac{c}{1-b}}{2}\left(2\right)\)

\(\Rightarrow\sqrt{\dfrac{ca}{b+ ca}}\le\dfrac{\dfrac{c}{1-a}+\dfrac{a}{1-c}}{2}\left(3\right)\)

\( \left(1\right)\left(2\right)\left(3\right)\Rightarrow A\le\dfrac{\dfrac{a}{1-b}+\dfrac{b}{1-a}+\dfrac{b}{1-c}+\dfrac{c}{1-b}+\dfrac{c}{1-a}+\dfrac{a}{1-c}}{2}=\dfrac{\dfrac{a+c}{1-b}+\dfrac{b+c}{1-a}+\dfrac{b+a}{1-c}}{2}=\dfrac{\dfrac{1-b}{1-b}+\dfrac{1-a}{1-a}+\dfrac{1-c}{1-c}}{2}=\dfrac{3}{2}\)

\(\Rightarrow A_{max}=\dfrac{3}{2}\Leftrightarrow a=b=c=\dfrac{1}{3}\)

8 tháng 1 2022

à e nhầm tìm giá trị lớn nhất ạ

26 tháng 12 2023

\(P=\dfrac{9}{ab+bc+ca}+\dfrac{2}{a^2+b^2+c^2}\)

\(=2\left[\dfrac{1}{a^2+b^2+c^2}+\dfrac{4}{2\left(ab+bc+ca\right)}\right]+\dfrac{5}{ab+bc+ca}\)

\(\ge2.\dfrac{\left(1+2\right)^2}{\left(a+b+c\right)^2}+\dfrac{5}{ab+bc+ca}\)

\(=\dfrac{18}{1}+\dfrac{5}{ab+bc+ca}\ge18+5.\dfrac{3}{\left(a+b+c\right)^2}=18+15=33\)

Đẳng thức xảy ra khi a=b=c=1/3.

Vậy GTNN của P là 33.

1 tháng 1

áp dụng bất đẳng thức phụ \(\dfrac{1}{a}+\dfrac{1}{b}\)\(\dfrac{4}{a+b}\)<=>(a-b)2≥0 (luôn đúng)
Ta có P≥\(\dfrac{\left(3+\sqrt{2}\right)^2}{\left(a+b+c\right)^2}\)=(3+\(\sqrt{2}\))2
Dấu = xảy ra <=> a=b=c=1/3