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Áp dụng bất đẳng thức Cô-si ta có:
\(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\sqrt[3]{\dfrac{a^2}{b^3}.\dfrac{1}{a}.\dfrac{1}{a}}=\dfrac{3}{b}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
Cộng theo vế ta được:
\(\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{a^2}{a^3}+\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{c^2}{a^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Xét \(\frac{a^3}{a^2+ab+b^2}-\frac{b^3}{a^2+ab+b^2}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=a-b\)
Tương tự, ta được: \(\frac{b^3}{b^2+bc+c^2}-\frac{c^3}{b^2+bc+c^2}=b-c\); \(\frac{c^3}{c^2+ca+a^2}-\frac{a^3}{c^2+ca+a^2}=c-a\)
Cộng theo vế của 3 đẳng thức trên, ta được: \(\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)\)\(-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)=0\)
\(\Rightarrow\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\)\(=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\)
Ta đi chứng minh BĐT phụ sau: \(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\)(*)
Thật vậy: (*)\(\Leftrightarrow\frac{2}{3}\left(a-b\right)^2\ge0\)*đúng*
\(\Rightarrow2LHS=\Sigma_{cyc}\frac{a^3+b^3}{a^2+ab+b^2}=\Sigma_{cyc}\text{ }\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}\)\(\ge\Sigma_{cyc}\text{ }\frac{\frac{1}{3}\left(a+b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=\frac{1}{3}\text{}\Sigma_{cyc}\left[\left(a+b\right)\right]=\frac{2\left(a+b+c\right)}{3}\)
\(\Rightarrow LHS\ge\frac{a+b+c}{3}=RHS\)(Q.E.D)
Đẳng thức xảy ra khi a = b = c
P/S: Có thể dùng BĐT phụ ở câu 3a để chứng minhxD:
1) ta chứng minh được \(\Sigma\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}=\Sigma\frac{b^4}{\left(a+b\right)\left(a^2+b^2\right)}\)
\(VT=\frac{1}{2}\Sigma\frac{a^4+b^4}{\left(a+b\right)\left(a^2+b^2\right)}\ge\frac{1}{4}\Sigma\frac{a^2+b^2}{a+b}\ge\frac{1}{8}\Sigma\left(a+b\right)=\frac{a+b+c+d}{4}\)
bài 2 xem có ghi nhầm ko
Áp dụng Svac
\(\Sigma\frac{a^3}{b+c}=\Sigma\frac{a^4}{ab+ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}=\frac{1}{2}\left(a^2+b^2+c^2\right)\)
"=" tại a=b=c
E thử làm cách khác ạ:))
Không mất tính tổng quát,giả sử \(a\ge b\ge c\)
\(\Rightarrow\hept{\begin{cases}a^2\ge b^2\ge c^2\\\frac{a}{b+c}\ge\frac{b}{a+c}\ge\frac{c}{a+b}\end{cases}}\)
Áp dụng BĐT Trebysev ta có:
\(a^2\cdot\frac{a}{b+c}+b^2\cdot\frac{b}{a+c}+c^2\cdot\frac{c}{a+b}\ge\frac{a^2+b^2+c^2}{3}\cdot\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
\(\ge\frac{a^2+b^2+c^2}{3}\cdot\frac{3}{2}\left(nesbitt\right)\)
\(=\frac{a^2+b^2+c^2}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
\(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}=\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\)
=> \(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)
Tương tự : \(\frac{b}{1+c^2}\ge b-\frac{bc}{2}\) ; \(\frac{c}{1+a^2}\ge c-\frac{ac}{2}\)
Cộng theo vế : \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge3-\frac{1}{2}\left(ab+bc+ac\right)\ge3-\frac{1}{2}.\frac{\left(a+b+c\right)^2}{3}=\frac{3}{2}\)
\(\Rightarrow\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge\frac{3}{2}\)
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\)
\(\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{a+c}+b+\frac{c^2}{a+b}+c\ge\frac{a+b+c}{2}+a+b+c\)
\(\Leftrightarrow a\left(\frac{a}{b+c}+1\right)+b\left(\frac{b}{a+c}+1\right)+c\left(\frac{c}{a+b}+1\right)\ge\frac{3}{2}\left(a+b+c\right)\)
\(\Leftrightarrow a\left(\frac{a+b+c}{b+c}\right)+b\left(\frac{a+b+c}{c+a}\right)+c\left(\frac{a+b+c}{a+b}\right)\ge\frac{3}{2}\left(a+b+c\right)\)
\(\Leftrightarrow\left(a+b+c\right)\frac{a}{b+c}+\left(a+b+c\right)\frac{b}{c+a}+\left(a+b+c\right)\frac{c}{a+b}\ge\frac{3}{2}\left(a+b+c\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\ge\frac{3}{2}\left(a+b+c\right)\)
\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\ge\frac{3}{2}+3\)
\(\Leftrightarrow\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\ge\frac{9}{2}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge\frac{9}{2}\)
\(\Leftrightarrow2\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge9\)
\(\Leftrightarrow\left(2a+2b+2c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge9\)
\(\Leftrightarrow\left(b+c+c+a+a+b\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge9\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}b+c+c+a+a+b\ge3\sqrt[3]{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\\\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\ge3\sqrt[3]{\frac{1}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}}\end{matrix}\right.\)
Nhân từng vế :
\(\Rightarrow\left(b+c+c+a+a+b\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge9\sqrt[3]{\left(b+c\right)\left(c+a\right)\left(a+b\right).\frac{1}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}}\)
\(\Rightarrow\left(b+c+c+a+a+b\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge9\left(đpcm\right)\)
Vậy với a ,b ,c > 0 thì \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\)
Áp dụng bất đẳng thức cô-si cho các số thực không âm ta có:
\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2}{b+c}\times\frac{b+c}{4}}=a\) (1)
\(\frac{b^2}{a+c}+\frac{a+c}{4}\ge2\sqrt{\frac{b^2}{a+c}\times\frac{a+c}{4}}=b\) (2)
\(\frac{c^2}{a+b}+\frac{a+b}{4}\ge2\sqrt{\frac{c^2}{a+b}\times\frac{a+b}{4}}=c\) (3)
Cộng (1),(2) và (3),vế theo vế ta được:
\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}+\frac{a+b+c}{2}\ge a+b+c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\) (đpcm)
Dấu "=" xảy ra khi :a=b=c
Vậy \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\) với a,b,c >0
Lời giải:
Áp dụng BĐT Bunhiacopxky ta có:
\(\left(\frac{a^4}{b^2+c^2}+\frac{b^4}{c^2+a^2}+\frac{c^4}{a^2+b^2}\right)\left(b^2+c^2+c^2+a^2+a^2+b^2\right)\ge\left(a^2+b^2+c^2\right)^2\)
\(\Leftrightarrow\frac{a^4}{b^2+c^2}+\frac{b^4}{a^2+c^2}+\frac{c^4}{a^2+b^2}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}=\frac{a^2+b^2+c^2}{2}\)
Ta có đpcm
Dấu bằng xảy ra khi \(\left|a\right|=\left|b\right|=\left|c\right|\)
mình nghĩ đề bài sai một chỗ :\(\frac{a^2}{b^2}\)chứ ko phải là \(\frac{a}{b^2}\)
khó quá chưa học