Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \frac{(a+b+c)^2}{b+c+a}=a+b+c\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$

Cách khác:

Áp dụng BĐT Cô-si cho các số dương ta có:

$\frac{a^2}{b}+b\geq 2a$

$\frac{b^2}{c}+c\geq 2b$

$\frac{c^2}{a}+a\geq 2c$

Cộng theo vế và thu gọn ta được:

$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq a+b+c$ (đpcm)

Dấu "=" xảy ra khi $a=b=c$

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)(ab+bc+ac)\geq (a+b+c)^2\)

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

Nhân theo vế 2 BĐT trên:

\(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2(ab+bc+ac).\frac{a+b+c}{abc}\geq [(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})]^2\)

\(\Leftrightarrow \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq [(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})]^2\)

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

Dấu "=" xảy ra khi $a=b=c$

\(\left(a+\frac{4b}{c^2}\right)\left(b+\frac{4c}{a^2}\right)\left(c+\frac{4a}{b^2}\right)\ge2\sqrt{\frac{4ab}{c^2}}.2\sqrt{\frac{4bc}{a^2}}.2\sqrt{\frac{4ac}{b^2}}=64\)

Dấu "=" xảy ra khi \(a=b=c=2\)

\(\frac{a^3}{b}+ab\ge2a^2\) ; \(\frac{b^3}{c}+bc\ge2b^2\); \(\frac{c^3}{a}+ac\ge2c^2\)

\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)\ge2\left(ab+bc+ca\right)-\left(ab+bc+ca\right)=ab+bc+ca\)

Dấu "=" xảy ra khi \(a=b=c\)

\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)=\frac{a+b+c}{2}\)

\(\Leftrightarrow\frac{a+b+c}{2}\ge\frac{3}{2}\)

\(\Leftrightarrow2\left(a+b+c\right)\ge6\)

\(\Leftrightarrow a+b+c\ge3\)

Dấu ''='' chỉ xảy ra khi \(a=b=c=1\left(đpcm\right)\)

THANKS

Vào thống kê hỏi đáp xem nhé. Bài này chỉ cần biểu diễn dưới dạng tổng bình phương là xong.

ta có \(\frac{a^3}{b^2+3}+\frac{b^3}{c^2+3}+\frac{c^3}{a^2+3}\ge\frac{3}{4}\) (***)

do ab+bc+ca=3 nên

VT (***)=\(\frac{a^3}{b^2+ab+bc+ca}+\frac{b^3}{c^2+ab+bc+ca}+\frac{c^3}{a^2+ab+bc+ca}\)

\(=\frac{a^3}{\left(b+c\right)\left(a+b\right)}+\frac{b^3}{\left(c+a\right)\left(b+c\right)}+\frac{c^3}{\left(a+b\right)\left(c+a\right)}\)

áp dụng bđt AM-GM ta có \(\frac{a^3}{\left(b+c\right)\left(c+a\right)}+\frac{b+c}{8}+\frac{a+b}{8}\ge\frac{3a}{4}\)

\(\Rightarrow\frac{a^3}{\left(b+c\right)\left(c+a\right)}\ge\frac{5a-2b-c}{8}\left(1\right)\)

chứng minh tương tự ta cũng được

\(\hept{\begin{cases}\frac{b^3}{\left(c+a\right)\left(a+b\right)}\ge\frac{5b-2c-a}{8}\left(2\right)\\\frac{c^3}{\left(a+b\right)\left(c+a\right)}\ge\frac{5c-2a-b}{8}\left(3\right)\end{cases}}\)

cộng theo vế với vế của (1),(2) và (3) ta được VT (***) \(\ge\frac{a+b+c}{4}\)

mặt khác ta dễ dàng chứng minh được \(a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}=3\)

đẳng thức xảy ra khi a=b=c=1 (đpcm)

ok cảm ơn bạn nha :)

Không mất tính tổng quát giả sử \(a\ge b\ge c>0\)

\(BĐT< =>\frac{a\left(b+c\right)\left(c+a\right)+b\left(a+b\right)\left(c+a\right)+c\left(a+b\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{3}{2}\)

\(< =>\frac{ac^2+ba^2+cb^2+\left(a+b+c\right)\left(ab+bc+ca\right)}{\left(a+b+c\right)\left(ab+bc+ca\right)-abc}\ge\frac{3}{2}\)

\(< =>2\left[ac^2+ba^2+cb^2+\left(a+b+c\right)\left(ab+bc+ca\right)\right]\ge3\left[\left(a+b+c\right)\left(...\right)-abc\right]\)

\(< =>2\left(ac^2+a^2b+cb^2\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-3abc\)

\(< =>ac^2+a^2b+cb^2\ge ca^2+ab^2+c^2b\)

\(< =>\left(c-b\right)\left(c-a\right)\left(a-b\right)\ge0\)(đúng)

Vậy ta có điều phải chứng minh

Ta có bất đẳng thức sau \(\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge9\)( cm = bunhia phân thức )

\(< =>1+\frac{a+b}{b+c}+\frac{a+b}{c+a}+1+\frac{b+c}{a+b}+\frac{b+c}{c+a}+1+\frac{c+a}{a+b}+\frac{c+a}{b+c}\ge9\)

\(< =>\frac{a}{a+b}+\frac{2a}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}+\frac{c}{b+c}+\frac{c}{c+a}\ge6\)(*)

Đặt \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\);\(B=\frac{a}{a+c}+\frac{b}{b+a}+\frac{c}{c+b}\);\(C=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

Khi đó bất đẳng thức (*) tương đương với \(A+B+2C\ge6\)

Do\(A+B=3\)\(=>2C\ge3=>C\ge\frac{3}{2}\)

Suy ra \(A+B+C\ge6-\frac{3}{2}=\frac{12-3}{2}=\frac{9}{2}\)(1)

Xét tổng :\(B+C=\frac{a}{a+c}+\frac{b}{b+a}+\frac{c}{c+b}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a+b}{a+c}+\frac{c+a}{b+c}+\frac{b+c}{a+b}\ge3\)(AM-GM) (2)

Từ (1) và (2) ta được \(A\ge\frac{9}{2}-3=\frac{3}{2}\)

Done !

Áp dụng BĐT AM-GM ta có: \(\frac{a}{b^3+ab}=\frac{1}{b}-\frac{b}{a+b^2}\ge\frac{1}{b}-\frac{b}{2\sqrt{ab^2}}=\frac{1}{b}-\frac{1}{2\sqrt{a}}\ge\frac{1}{b}-\frac{1}{4}\left(\frac{1}{a}+1\right)\)

Tương tự có: \(\hept{\begin{cases}\frac{b}{c^3+ca}\ge\frac{1}{c}-\frac{1}{4}\left(\frac{1}{b}+1\right)\\\frac{c}{a^3+ca}\ge\frac{1}{a}-\frac{1}{4}\left(\frac{1}{c}+1\right)\end{cases}}\)

Cộng 3 vế BĐT ta được:  \(\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ca}\ge\frac{3}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\)

Bài toán quy về chứng minh \(\frac{3}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\Leftrightarrow\left(\frac{1}{a}+a\right)\left(\frac{1}{b}+b\right)\left(\frac{1}{c}+c\right)\ge3+a+b+c=6\)

BĐT cuối hiển nhiên đúng vì theo BĐT AM-GM ta có:

\(\hept{\begin{cases}\frac{1}{a}+a\ge2\\\frac{1}{b}+b\ge2\\\frac{1}{c}+c\ge2\end{cases}}\)

Dấu "=" xảy ra <=> a=b=c=1

\(\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ac}\)

\(=\frac{a}{b\left(b^2+a\right)}+\frac{b}{c\left(c^2+b\right)}+\frac{c}{a\left(a^2+c\right)}\)

\(=\frac{1}{b}-\frac{b}{b^2+a}+\frac{1}{c}-\frac{c}{c^2+b}+\frac{1}{a}-\frac{a}{a^2+c}\)

\(\ge\frac{1}{b}-\frac{b}{2b\sqrt{a}}+\frac{1}{c}-\frac{c}{2c\sqrt{b}}+\frac{1}{a}-\frac{a}{2a\sqrt{c}}\)

\(=\frac{3}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{4}\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\frac{1}{4}\left(\frac{1}{b}-\frac{2}{\sqrt{b}}+1\right)+\frac{1}{4}\left(\frac{1}{c}-\frac{1}{\sqrt{c}}+1\right)\)\(-\frac{3}{4}\)

\(\ge\frac{3}{4}.\frac{9}{a+b+c}+\frac{1}{4}\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{1}{4}\left(\frac{1}{\sqrt{b}}-1\right)^2+\frac{1}{4}\left(\frac{1}{\sqrt{b}}-1\right)^2-\frac{3}{4}\)

\(\ge\frac{3}{2}\)

Dấu "=" xảy ra <=> a = b = c = 1.

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

=>  \(A+3=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\)

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

\(=\frac{1}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

\(\ge\frac{1}{2}.3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}.3\sqrt[3]{\frac{1}{a+b}.\frac{1}{b+c}.\frac{1}{c+a}}=\frac{9}{2}\)   (AM - GM)

=>  \(A\ge\frac{9}{2}-3=\frac{3}{2}\)  (đpcm)

Đặt \(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(A=\frac{a^2}{ba+ca}+\frac{b^2}{cb+ba}+\frac{c^2}{ac+bc}\)

Áp dụng BĐT Cauchy-schwarz ta có: 

\(A=\frac{a^2}{ba+ca}+\frac{b^2}{cb+ba}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2.\left(ab+bc+ca\right)}\)

Ta c/m BĐT phụ \(ab+bc+ca\le\frac{1}{3}.\left(a+b+c\right)^2\)( tự c/m)

Áp dụng: 

\(A\ge\frac{\left(a+b+c\right)^2}{2.\frac{1}{3}\left(a+b+c\right)^2}=\frac{1}{\frac{2}{3}}=\frac{3}{2}\)

                                                 đpcm

Tham khảo nhé~

P = a/(b+c) + b/(c+a) + c/(a+b) 
P + 3 = 1+ a/(b+c) + 1+ b/(c+a) + 1+ c/(a+b) 
P + 3 = (a+b+c)/(b+c) + (a+b+c)/(b+c) + (a+b+c)/(c+a) 
P + 3 = (a+b+c)[1/(b+c) + 1/(c+a) + 1/(a+b)] (*) 

ad bđt cô si cho 3 số: 
2(a+b+c) = (a+b) + (b+c) + (c+a) ≥ 3.³√(a+b)(b+c)(c+a) 
1/(b+c) + 1/(c+a) + 1/(a+b) ≥ 3.³√1/(a+b)(b+c)(c+a) 

nhân lại vế theo vế 2 bđt: 2(a+b+c)[1/(b+c) + 1/(c+a) + 1/(a+b)] ≥ 9 
=> P + 3 ≥ 9/2 => P ≥ 3/2 (đpcm) ; dấu "=" khi a = b = c 
- - - 
cách khác: P = a/(b+c) + b/(c+a) + c/(a+b) 
M = b/(b+c) + c/(c+a) + a/(a+b) 
N = c/(b+c) + a/(c+a) + b/(a+b) 

Thấy: M + N = 3 
P + M = (a+b)/(b+c) + (b+c)/(c+a) + (c+a)/(a+b) ≥ 3 (cô si cho 3 số) 
P + N = (a+c)/(b+c) + (b+a)/(c+a) + (c+b)/(a+b) ≥ 3 (cô si) 

=> 2P + M + N ≥ 6 => 2P + 3 ≥ 6 => P ≥ 3/2 (đpcm) ; đẳng thức khi a = b = c