Bài toán tổng quát: Đề này n lẻ mới đúng nhé

Ta có:

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

\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\)

\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{ac+bc+c^2}\right)=0\)

\(\Leftrightarrow\dfrac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{ab\left(ac+bc+c^2\right)}=0\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)

Nếu \(a=-b\Rightarrow a^n=-b^n\) và \(\dfrac{1}{a^n}=\dfrac{-1}{b^n}\)

Ta có: \(\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{c^n}\)

\(\dfrac{1}{a^n+b^n+c^n}=\dfrac{1}{c^n}\)

VT = VP => ĐPCM

Còn ý còn lại thì dựa trên bài này mà biến đổi một tí là ra

Bài 1: 

a) Áp dụng bđt Cô - si:

\(\dfrac{a}{b^2}+\dfrac{1}{a}\ge\dfrac{2}{b}\)

Tương tự với 2 phân thức còn lại của vế trái rồi cộng lại, ta có:

\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\)

=> đpcm

Bài dù a + b + c = 2021 hay 1 số bất kì thì bđt luôn \(\ge\dfrac{3}{2}\). Bạn có thể tham khảo bđt Nesbitt

Bài 2:

\(P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

\(=\dfrac{2021-\left(b+c\right)}{b+c}+\dfrac{2021-\left(c+a\right)}{c+a}+\dfrac{2021-\left(a+b\right)}{a+b}\)

\(=2021\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)-3\)

Áp dụng BĐT Svacxo, ta có

\(P\) ≥ \(\dfrac{9}{2}-3=\dfrac{3}{2}\)

Dấu"=" ⇔ ...

Sau khi đã đi tham khảo 7749 người thì đã cho ra một kết quả:v

Bài 2. \(P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

\(P=\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1+\dfrac{c}{a+b}+1-3\)

\(P=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\)

\(P=\dfrac{(2a+2b+3c)( \dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b})}{2}-3 ≥ \dfrac{9}{2}-3=\dfrac{3}{2}\)

Dấu `"="` xảy ra:

\(\Leftrightarrow \begin{cases} a=b=c\\ a+b+c=2021 \end{cases} \)

\(\Leftrightarrow a=b=c=\dfrac{2021}{3}\)

Vậy \(min \) \(P=\dfrac{3}{2}\) khi \(a=b=c=\dfrac{2021}{3}\)

Lời giải:

Ta có:

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

\(\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)

\(\Leftrightarrow \frac{a+b}{ab}+\frac{a+b}{c(a+b+c)}=0\)

\(\Leftrightarrow (a+b)\left(\frac{1}{ab}+\frac{1}{c(a+b+c)}\right)=0\)

\(\Leftrightarrow \frac{(a+b)[c(a+b+c)+ab]}{abc(a+b+c)}=0\)

\(\Leftrightarrow (a+b)(b+c)(c+a)=0\)

Xét : \(A=\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}-\frac{1}{a^n+b^n+c^n}\)

\(A=\frac{a^n+b^n}{a^nb^n}+\frac{a^n+b^n}{c^n(a^n+b^n+c^n)}\)

\(A=(a^n+b^n)\left(\frac{1}{a^nb^n}+\frac{1}{c^n(a^n+b^n+c^n)}\right)\)

\(A=\frac{(a^n+b^n)[c^n(a^n+b^n+c^n)+a^nb^n]}{a^nb^nc^n(a^n+b^n+c^n)}\)

\(A=\frac{(a^n+b^n)(b^n+c^n)(c^n+a^n)}{a^nb^nc^n(a^n+b^n+c^n)}\)

Vì $n$ lẻ nên :

\((a^n+b^n)(b^n+c^n)(c^n+a^n)=(a+b)(b+c)(c+a)(a^{n-1}+....+b^{n-1})(b^{n-1}+..+c^{n-1})(c^{n-1}+...+a^{n-1})\)

\(=0\) do \((a+b)(b+c)(c+a)=0\)

Do đó: \(A=0\Leftrightarrow \frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}\)

Với cả 3 phần thì dấu "=" xảy ra tại a=b=c=1.

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

(Cosi) \(\ge a-\dfrac{ab^2}{2b}=a-\dfrac{ab}{2}\)

Tương tự : \(\dfrac{b}{1+c^2}\ge b-\dfrac{bc}{2};\dfrac{c}{1+a^2}\ge c-\dfrac{ca}{2}\)

\(\Rightarrow P\ge\left(a+b+c\right)-\dfrac{ab+bc+ca}{2}\ge\left(CS\right)\left(a+b+c\right)-\dfrac{\left(a+b+c\right)^2}{6}=3-\dfrac{3^2}{6}=\dfrac{3}{2}\)

b) \(\dfrac{1}{a^2+1}=1-\dfrac{a^2}{a^2+1}\ge\left(CS\right)1-\dfrac{a^2}{2a}=1-\dfrac{a}{2}\)

Tương tự : \(\dfrac{1}{b^2+1}\ge1-\dfrac{b}{2};\dfrac{1}{c^2+1}\ge1-\dfrac{c}{2}\)

\(\Rightarrow P\ge3-\dfrac{a+b+c}{2}=3-\dfrac{3}{2}=\dfrac{3}{2}\)

c)\(P=\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}=\left(\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\right)+\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\ge\dfrac{3}{2}+\dfrac{3}{2}=3\)

@Akai Haruma

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\Leftrightarrow\left(ab+ac+bc\right)\left(a+b+c\right)=abc\Leftrightarrow a^2b+ab^2+abc+ac^2+abc+ac^2+abc+b^2c+bc^2=abc\Leftrightarrow a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abc=0\Leftrightarrow\left(a^2b+ab^2\right)+\left(a^2c+abc\right)+\left(b^2c+abc\right)+\left(ac^2+bc^2\right)=0\Leftrightarrow ab\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)=0\Leftrightarrow\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\Leftrightarrow\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=0\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a+b=0\\b+c=0\\c+a=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)

TH1:a=-b

\(\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n}-\dfrac{1}{a^n}+\dfrac{1}{c^n}=\dfrac{1}{c^n}\)(vì n lẻ)

\(\dfrac{1}{a^n+b^n+c^n}=\dfrac{1}{a^n-a^n+c^n}=\dfrac{1}{c^n}\)(vì n lẻ)

Suy ra \(\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n+b^n+c^n}\)

Chứng minh tương tự trong các trường hợp b=-c và c=-a

Vậy \(\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n+b^n+c^n}\)

Bài này phải thêm dữ kiện n lẻ mình mới làm được

Áp dụng bđt Svácxơ, ta có:

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

\(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

Áp dụng, thay vào A, ta có: 

\(A\le\text{Σ}\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

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

Dấu "="⇔\(a=b=c=1\)

= chịu

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