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Ta có : \(\hept{\begin{cases}\frac{a^3}{a^2+b^2+ab}=\frac{a^4}{a\left(a^2+b^2+ab\right)}=\frac{a^4}{a^3+ab^2+a^2b}=\frac{a^4}{a^3+ab\left(a+b\right)}\\\frac{b^3}{b^2+c^2+bc}=\frac{b^4}{b\left(b^2+c^2+bc\right)}=\frac{b^4}{b^3+bc^2+b^2c}=\frac{b^4}{b^3+bc\left(b+c\right)}\\\frac{c^3}{c^2+a^2+ca}=\frac{c^4}{c\left(c^2+a^2+ca\right)}=\frac{c^4}{c^3+ca^2+c^2a}=\frac{c^4}{c^3+ca\left(c+a\right)}\end{cases}}\)
Khi đó bất đẳng thức được viết lại thành :
\(\frac{a^4}{a^3+ab\left(a+b\right)}+\frac{b^4}{b^3+bc\left(b+c\right)}+\frac{c^4}{c^3+ca\left(c+a\right)}\ge\frac{a+b+c}{3}\)
Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :
\(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^3+b^3+c^3+ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}\)
Dễ dàng phân tích \(a^3+b^3+c^3+ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)=\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)
=> \(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}=\frac{a^2+b^2+c^2}{a+b+c}\)
Xét bất đẳng thức phụ : 3( a2 + b2 + c2 ) ≥ ( a + b + c )2
<=> 3a2 + 3b2 + 3c2 - a2 - b2 - c2 - 2ab - 2bc - 2ca ≥ 0
<=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca ≥ 0
<=> ( a - b )2 + ( b - c )2 + ( c - a )2 ≥ 0 ( đúng )
Khi đó áp dụng vào bài toán ta có : \(VT\ge\frac{a^2+b^2+c^2}{a+b+c}=\frac{\frac{\left(a+b+c\right)^2}{3}}{a+b+c}=\frac{a+b+c}{3}\)( đpcm )
Đẳng thức xảy ra <=> a=b=c
bài này mới được thầy sửa hồi chiều nè @@
Vì a,b dương => ( a + b ) ( a - b )2 \(\ge\)0 => a3 + b3 \(\ge\)ab ( a + b )
BĐT tương đương với 3a3\(\ge\)2a3 + 2ab ( a + b ) - b3 = 2a3 + 2a2b + 2ab2 - a2b - ab2 - b3 = ( a2 + ab + b3 ) ( 2a - b )
Suy ra : \(\frac{a^3}{a^2+ab+b^2}\ge\frac{2a-b}{3}\)(1)
Chứng minh tương tự ta được : \(\frac{b^3}{b^2+bc+c^2}\ge\frac{2b-c}{3}\)(2) ; \(\frac{c^3}{c^2+ca+a^2}\ge\frac{2c-a}{3}\)(3)
Từ (1) ; (2) và (3) => \(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\ge\frac{a+b+c}{3}\)(đpcm)
\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\ge\frac{a+b+c}{3}\)
\(\Leftrightarrow\frac{a^4}{a^3+a^2b+ab^2}+\frac{b^4}{b^3+b^2c+bc^2}+\frac{c^4}{c^3+c^2a+a^2c}\ge\frac{a+b+c}{3}\)
\(\Leftrightarrow\frac{\left(a^2\right)^2}{a^3+a^2b+ab^2}+\frac{\left(b^2\right)^2}{b^3+b^2c+bc^2}+\frac{\left(c^2\right)^2}{c^3+c^2a+a^2c}\ge\frac{a+b+c}{3}\)
Áp dụng bất đẳng thức cộng mẫu số cho vế trái
\(\Rightarrow\frac{\left(a^2\right)^2}{a^3+a^2b+ab^2}+\frac{\left(b^2\right)^2}{b^3+b^2c+bc^2}+\frac{\left(c^2\right)^2}{c^3+c^2a+a^2c}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^3+b^3+c^3+a^2b+ab^2+b^2c+bc^2+c^2a+a^2c}\)
\(\Rightarrow\frac{\left(a^2\right)^2}{a^3+a^2b+ab^2}+\frac{\left(b^2\right)^2}{b^3+b^2c+bc^2}+\frac{\left(c^2\right)^2}{c^3+c^2a+a^2c}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^3+a^2b+a^2c\right)+\left(b^3+b^2c+ab^2\right)+\left(c^3+c^2a+bc^2\right)}\)
\(\Rightarrow\frac{\left(a^2\right)^2}{a^3+a^2b+ab^2}+\frac{\left(b^2\right)^2}{b^3+b^2c+bc^2}+\frac{\left(c^2\right)^2}{c^3+c^2a+a^2c}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2\left(a+b+c\right)+b^2\left(a+b+c\right)+c^2\left(a+b+c\right)}\)
\(\Rightarrow\frac{\left(a^2\right)^2}{a^3+a^2b+ab^2}+\frac{\left(b^2\right)^2}{b^3+b^2c+bc^2}+\frac{\left(c^2\right)^2}{c^3+c^2a+a^2c}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)\left(a+b+c\right)}=\frac{a^2+b^2+c^2}{a+b+c}\)
Chứng minh rằng: \(\frac{a^2+b^2+c^2}{a+b+c}\ge\frac{a+b+c}{3}\)
\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
Áp dụng bất đẳng thức Bunhiacopski cho 3 bộ số thực không âm
\(\Rightarrow3\left(a^2+b^2+c^2\right)=\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)( đpcm )
Vậy \(\frac{a^2+b^2+c^2}{a+b+c}\ge\frac{a+b+c}{3}\)
Vì \(\frac{\left(a^2\right)^2}{a^3+a^2b+ab^2}+\frac{\left(b^2\right)^2}{b^3+b^2c+bc^2}+\frac{\left(c^2\right)^2}{c^3+c^2a+a^2c}\ge\frac{a^2+b^2+c^2}{a+b+c}\)
\(\Rightarrow\frac{\left(a^2\right)^2}{a^3+a^2b+ab^2}+\frac{\left(b^2\right)^2}{b^3+b^2c+bc^2}+\frac{\left(c^2\right)^2}{c^3+c^2a+a^2c}\ge\frac{a+b+c}{3}\)
\(\Leftrightarrow\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\ge\frac{a+b+c}{3}\) ( đpcm )
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)
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.
\(A=\frac{\frac{1}{2}a^2\left(\sqrt[3]{b}+\sqrt[3]{c}+1\right)\left[\left(\sqrt[3]{b}-\sqrt[3]{c}\right)^2+\left(\sqrt[3]{b}-1\right)^2+\left(\sqrt[3]{c}-1\right)^2\right]}{2\left(a+2\right)\left(a+\sqrt[3]{bc}\right)}\ge0\)
\(\Sigma_{cyc}\frac{a^2}{a+\sqrt[3]{bc}}=\Sigma_{cyc}A+\Sigma_{cyc}\frac{2\left(a-1\right)^2}{3\left(a+2\right)}+\frac{5}{6}\left(a+b+c\right)-1\ge\frac{5}{6}\left(a+b+c\right)-1=\frac{3}{2}\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\frac{a^2}{a+\sqrt[3]{bc}}+\frac{b^2}{b+\sqrt[3]{ca}}+\frac{c^2}{c+\sqrt[3]{ab}}\)\(\ge\frac{\left(a+b+c\right)^2}{a+b+c+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
\(\Rightarrow\frac{a^2}{a+\sqrt[3]{bc}}+\frac{b^2}{b+\sqrt[3]{ca}}+\frac{c^2}{c+\sqrt[3]{ab}}\)\(\ge\frac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
Chứng minh rằng : \(\frac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\frac{3}{2}\)
\(\Leftrightarrow18\ge3\left(3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}\right)\)
\(\Leftrightarrow18\ge9+3\sqrt[3]{bc}+3\sqrt[3]{ca}+3\sqrt[3]{ab}\)
\(\Leftrightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)
Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm
\(\Rightarrow\hept{\begin{cases}a+b+1\ge3\sqrt[3]{ab}\\b+c+1\ge3\sqrt[3]{bc}\\c+a+1\ge3\sqrt[3]{ca}\end{cases}}\)
\(\Rightarrow2\left(a+b+c\right)+3\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)
\(\Rightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\left(đpcm\right)\)
Vì \(\frac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\frac{3}{2}\)
Mà \(\frac{a^2}{a+\sqrt[3]{bc}}+\frac{b^2}{b+\sqrt[3]{ca}}+\frac{c^2}{c+\sqrt[3]{ab}}\ge\frac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
\(\Rightarrow\frac{a^2}{a+\sqrt[3]{bc}}+\frac{b^2}{b+\sqrt[3]{ca}}+\frac{c^2}{c+\sqrt[3]{ab}}\ge\frac{3}{2}\left(đpcm\right)\)
Chúc bạn học tốt !!!
Áp dụng BĐT Cauchy Shwarz dạng Engel và BĐT AM - GM, ta có:
\(\frac{a^5}{bc}+\frac{b^5}{ac}+\frac{c^5}{ab}\)
\(=\frac{a^6}{abc}+\frac{b^6}{abc}+\frac{c^6}{abc}\)
\(\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}\)
\(\ge\frac{\left(a^3+b^3+c^3\right)^2}{a^3+b^3+c^3}\)
\(=a^3+b^3+c^3\left(\text{đ}pcm\right)\)
Dấu "=" xảy ra khi a = b = c
\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}=\frac{1}{abc}\left(a^6+b^6+c^6\right)\)
\(\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}\ge\frac{3abc\left(a^3+b^3+c^3\right)}{3abc}=a^3+b^3+c^3\)
CM BĐT : \(\left(x^2+y^2+z^2\right)^2\ge3\left(x^3y+y^3z+z^3x\right)\) ( * )
\(\frac{a}{ab+1}=\frac{a\left(ab+1\right)-a^2b}{ab+1}=a-\frac{a^2b}{ab+1}\)
TT ....
Áp dụng BĐT ( * ) với x = \(\sqrt{a}\); y = \(\sqrt{b}\); z = \(\sqrt{c}\) vào bài toán, ta có :
\(\frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}=a+b+c-\frac{a^2b}{ab+1}-\frac{b^2c}{bc+1}-\frac{c^2a}{ac+1}\)
\(\ge3-\frac{a^2b}{2\sqrt{ab}}-\frac{b^2c}{2\sqrt{bc}}-\frac{c^2a}{2\sqrt{ac}}=3-\frac{\sqrt{a^3b}+\sqrt{b^3c}+\sqrt{c^3a}}{2}\ge3-\frac{\frac{\left(a+b+c\right)^2}{3}}{2}=\frac{3}{2}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
\(\frac{a^3}{bc}+b+c\ge3\sqrt[3]{\frac{a^3.b.c}{b.c}}=3a\)
\(\frac{b^3}{ca}+a+c\ge3b\)
\(\frac{c^3}{ab}+a+b\ge3c\)
\(vt+2\left(a+b+c\right)\ge3\left(a+b+c\right)\)
\(vt\ge a+b+c\)