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\(VT=\frac{a^3}{a^2+abc}+\frac{b^3}{b^2+abc}+\frac{c^3}{c^2+abc}\)
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 !!!
Ta có:\(\sqrt{\frac{bc}{a+bc}}=\sqrt{\frac{bc}{a\left(a+b\right)+c\left(a+b\right)}}\)
\(=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\) (Áp dụng BĐT AM-GM)
Tương tự với hai BĐT còn lại và cộng theo vế ta thu được đpcm.
Từ giả thiết ta có:
\(\left(a+b+c\right)^3=a^2+b^2+c^2+2\left(ab+bc+ca\right)=1\)
\(\frac{3}{ab+bc+ac}=\frac{3\left(a^2+b^2+c^2\right)+6\left(ab+bc+ca\right)}{ab+bc+c}=\frac{3\left(a^2+b^2+c^2\right)}{ab+bc+ca}+6\)
\(\frac{2}{a^2+b^2+c^2}=\frac{2\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)}{a^2+b^2+c^2}=2+\frac{4\left(ab+bc+ca\right)}{a^2+b^2+c^2}\)
Áp dụng bđt Cosi cho 2 số dương ta có:
\(\frac{3}{ab+bc+ca}+\frac{2}{a^2+b^2+c^2}\ge6+2+2\sqrt{\frac{3\left(a^2+b^2+c^2\right)4\left(ab+bc+ca\right)}{\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}}=8+2\sqrt{12}\)
\(>8+2\sqrt{9}=14\)
\(A=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\left(a^3+b^3+c^3\right)\frac{1}{abc}\)
Cm với a+b+c=0 thì \(a^3+b^3+c^3=3abc\)(1) .Từ đó tính dc A, muốn cm(1) bạn xét hiệu nhé
\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)(luôn đúng vì a+b+c=0)
\(VT=\frac{a^3}{a^2+abc}+\frac{b^3}{b^2+abc}+\frac{c^3}{c^2+abc}\)
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 !!!
Áp dụng BĐT Bunhiacopxki, ta có:
\(\left(a+b+c\right)\left(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right)\ge\left(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\right)^2\)
Mà \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{bc}{1+bc+1}=1\)
\(\Rightarrow\left(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right)\left(a+b+c\right)\ge1\)
\(\Rightarrow\frac{a}{\left(ab+b+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\ge\frac{1}{a+b+c}\)
\(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\ge\frac{1}{a+b+c}\)
ta có \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}=1\)
đặt \(H=\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\)
áp dụng bất đẳng thức bunhiacopxki ta có
\(H\left(a+b+c\right)\ge\left(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\right)^2=1\)
\(\Rightarrow H\ge\frac{1}{a+b+c}\)
hay \(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\ge\frac{1}{a+b+c}\)
Áp dụng : x + y + z = 0 suy ra x3 + y3 + z3 = 3xyz
1/a + 1/2b + 1/3c = 0 = >... rồi biến đổi nhé
Ta có : \(\frac{a^2-bc}{a}+\frac{b^2-ac}{b}+\frac{c^2-ab}{c}=0\)
=> \(a-\frac{bc}{a}+b-\frac{ac}{b}+c-\frac{ab}{c}=0\)
=> \(a+b+c=\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\)
=> \(a+b+c=abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
=> \(\frac{a+b+c}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
=> \(\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
=> \(\frac{2}{bc}+\frac{2}{ac}+\frac{2}{ab}=\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}\)
=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{bc}-\frac{2}{ac}-\frac{2}{ac}=0\)
=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{a^2}-\frac{2}{ac}+\frac{1}{c^2}\right)+\left(\frac{1}{b^2}-\frac{1}{bc}+\frac{1}{c^2}\right)=0\)
=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{a}-\frac{1}{c}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2=0\)
=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{a}-\frac{1}{c}=0\\\frac{1}{b}-\frac{1}{c}=0\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{a}=\frac{1}{b}\\\frac{1}{a}=\frac{1}{c}\\\frac{1}{b}=\frac{1}{c}\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\left(\text{đpcm}\right)\)