Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng : x + y + z = 0 suy ra x3 + y3 + z3 = 3xyz
1/a + 1/2b + 1/3c = 0 = >... rồi biến đổi nhé
\(VT=\frac{a^3}{a^2+abc}+\frac{b^3}{b^2+abc}+\frac{c^3}{c^2+abc}\)
Xét \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)
\(\Rightarrow VT=\frac{a^3}{a^2+ab+bc+ac}+\frac{b^3}{b^2+ab+bc+ac}+\frac{c^3}{c^2+ab+bc+ac}\)
\(\Leftrightarrow VT=\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{b^3}{\left(b+a\right)\left(b+c\right)}+\frac{c^3}{\left(c+b\right)\left(c+a\right)}\)
Áp dụng bdt Cauchy ta có :
\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)
Thiết lập tương tự và thu lại ta có :
\(VT+\frac{a+b+c}{2}\ge\frac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\frac{3}{4}\left(a+b+c\right)--\frac{1}{2}\left(a+b+c\right)=\frac{a+b+c}{4}\left(đpcm\right)\)
Dấu " = " xảy ra khi \(a=b=c=3\)
Chúc bạn học tốt !!!
\(VT=\frac{a^3}{a^2+abc}+\frac{b^3}{b^2+abc}+\frac{c^3}{c^2+abc}\)
Xét \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)
\(\Rightarrow VT=\frac{a^3}{a^2+ab+bc+ac}+\frac{b^3}{b^2+ab+bc+ac}+\frac{c^3}{c^2+ab+bc+ac}\)
\(\Leftrightarrow VT=\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{b^3}{\left(b+a\right)\left(b+c\right)}+\frac{c^3}{\left(c+b\right)\left(c+a\right)}\)
Áp dụng BĐT Cauchy ta có :
\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)
Thiết lập tương tự và thu lại ta có :
\(VT+\frac{a+b+c}{2}\ge\frac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{2}\left(a+b+c\right)=\frac{a+b+c}{4}\left(đpcm\right)\)
Dấu " = " xảy ra khi \(a=b=c=3\)
Chúc bạn học tốt !!!
\(P=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3} +\frac{abc}{b^3}\)
\(=abc.\left(\frac{1}{c^3}+\frac{1}{a^3}+\frac{1}{b^3}\right)\)Mà nếu \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
thì \(\frac{1}{c^3}+\frac{1}{a^3}+\frac{1}{b^3}=\frac{3}{abc}\)\(\Rightarrow P=abc.\frac{3}{abc}=3\)
Ta có :
\(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(=\frac{abc.3}{\left(abc\right)}=3\)
Cách I HĐT a³ + b³ + c³ = [(a + b + c)(a² + b²+ c²-ab-bc-ca)+3abc]
Ta có:
bc/a²+ac/b²+ ab/c²=abc/a³+abc/b³+abc/c³
=abc(1/a³ + 1/b³ + 1/c³)
=abc[(1/a + 1/b + 1/c)(1/a² + 1/b²+ 1/c²-1/ab-1/bc-1/ca)+3/abc](áp dụng HĐt trên)
=abc.3/(abc)=3
Cách II:
Từ giả thiết suy ra:
(1/a +1/b)³=-1/c³
=>1/a³+1/b³+1/c³=-3.1/a.1/b(1/a+1/b)=3...
=>bc/a²+ac/b²+ ab/c²=abc/a³+abc/b³+abc/c³
=abc(1/a³ + 1/b³ + 1/c³)
=abc.3/(abc)=3
Áp dụng bđt Cô-si: \(\frac{a}{bc}+\frac{b}{ac}\ge2\sqrt{\frac{a}{bc}.\frac{b}{ac}}=\frac{2}{c}\)
\(\frac{b}{ac}+\frac{c}{ab}\ge2\sqrt{\frac{b}{ac}.\frac{c}{ab}}=\frac{1}{a}\)
\(\frac{c}{ab}+\frac{a}{bc}\ge2\sqrt{\frac{c}{ab}.\frac{a}{bc}}=\frac{1}{b}\)
cộng vế với vế ta được \(2\left(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
=>\(A=\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2}\)
Dấu "=" xảy ra khi a=b=c=2
Vậy minA=3/2 khi a=b=c=2
\(VT=\frac{a^3}{a^2+abc}+\frac{b^3}{b^2+abc}+\frac{c^3}{c^2+abc}\)
Xét \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)
\(\Rightarrow VT=\frac{a^3}{a^2+ab+bc+ac}+\frac{b^3}{b^2+ab+bc+ac}+\frac{c^3}{c^2+ab+bc+ac}\)
\(\Leftrightarrow VT=\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{b^3}{\left(b+a\right)\left(b+c\right)}+\frac{c^3}{\left(c+b\right)\left(c+a\right)}\)
Áp dụng bất đẳng thức Cauchy ta có :
\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)
Thiết lập tương tự và thu lại ta có :
\(VT+\frac{a+b+c}{2}\ge\frac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{2}\left(a+b+c\right)=\frac{a+b+c}{4}\) ( đpcm)
Dấu " = " xảy ra khi \(a=b=c=3\)
Chúc bạn học tốt !!!
P= abc(\(\frac{1}{^{a^3}}\)+\(\frac{1}{b^3}\)+\(\frac{1}{c^3}\)) = abc[(\(\frac{1}{a}\)+\(\frac{1}{b}\))3+\(\frac{1}{c^3}\)-\(\frac{3}{a^2b}\)-\(\frac{3}{ab^2}\)]=abc[(\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\))(....)- \(\frac{3}{a^2b}\)-\(\frac{3}{ab^2}\)]
=abc.(- \(\frac{3}{a^2b}\)-\(\frac{3}{ab^2}\)) =-3(\(\frac{c}{a}\)+\(\frac{c}{b}\)) = -3c(\(\frac{1}{a}\)+\(\frac{1}{b}\)) = -3c.\(\frac{-1}{c}\)=3
P = 3
Đầu tiên,bạn cần chứng minh x + y + z = 0 thì x3 + y3 + z3 = 3xyz ( Bạn ko biết c/m thì hỏi nhé)
Thay\(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3.\frac{1}{a}.\frac{1}{b}.\frac{1}{c}=\frac{3}{abc}\)
\(\Rightarrow M=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}=abc\left(\frac{1}{c^3}+\frac{1}{a^3}+\frac{1}{b^3}\right)=abc.\frac{3}{abc}=3\)