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\(\frac{bc+a^2}{a+b}+\frac{ac+b^2}{b+c}+\frac{ab+c^2}{a+c}\ge\)a+b+c
<=>\(\frac{bc+a^2}{a+b}-a+\frac{ac+b^2}{b+c}-b+\frac{ab+c^2}{a+c}-c\ge0\)
<=>\(\frac{b\left(c-a\right)}{a+b}+\frac{c\left(a-b\right)}{b+c}+\frac{a\left(b-c\right)}{a+c}\ge0\)
<=>\(\frac{b\left(b+c\right)\left(a+c\right)\left(a-c\right)}{\left(a+b\right)\left(c+c\right)\left(a+c\right)}\)+\(\frac{c\left(a+c\right)\left(a-b\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)+\(\frac{a\left(a+b\right)\left(b-c\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
<=>\(\frac{b^2c^2-b^2a^2+bc^3-a^2bc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)+\(\frac{a^3c-ab^2c+c^2a^2-b^2c^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)+\(\frac{a^2b^2-a^2c^2+ab^3-abc^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
<=>\(\frac{bc^3+a^3c+ab^3-a^2bc-ab^2c-abc^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
<=>\(\frac{2bc^3+2a^3c+2ab^3-2a^2bc-2ab^2c-2abc^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)>=0
<=>\(\frac{bc\left(c-a\right)^2+ac\left(a-b\right)^2+ab\left(b-c\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)(đung voi moi a,b,c >0)
Dấu ''='' xay ra khi a=b=c
1.
\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)
\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
Dấu "=" khi \(a=b=c\)
2.
\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)
Dấu "=" khi \(a=b=c=d\)
\(\frac{a^3}{bc}+\frac{b^3}{ca}=\frac{a^4}{abc}+\frac{b^4}{abc}\ge\frac{\left(a^2+b^2\right)^2}{2abc}\ge\frac{2ab\left(a^2+b^2\right)}{2abc}=\frac{a^2+b^2}{c}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b\)
viết các bđt tương tự rồi cộng vế theo vế là được
a/Xét hiệu ta có: \(\frac{a^3}{b}+\frac{b^3}{b}-a^2-ab=\left(a+b\right)\left(\frac{a^2-ab+b^2}{b}\right)-a\left(a+b\right)\)
\(=\left(a+b\right)\left(\frac{a^2}{b}-2a+b\right)=\left(a+b\right)\left(\frac{a}{\sqrt{b}}+\sqrt{b}\right)^2\ge0\)
\(\RightarrowĐPCM\)
b/Tương tự ở câu a, ta cũng có:
\(\frac{a^3}{b}\ge a^2+ab-b^2\left(1\right),\frac{b^3}{c}\ge b^2+bc-c^2\left(2\right),\frac{c^3}{a}\ge c^2+ca-a^2\left(3\right)\)
Cộng (1),(2) và (3) \(VT\ge a^2+ab-b^2+b^2+bc-c^2+C^2+bc-a^2=ab+bc+ca\left(ĐPCM\right)\)
a) Ta có: \(\frac{a^2}{a+b}-\frac{b^2}{a+b}+\frac{b^2}{b+c}-\frac{c^2}{b+c}+\frac{c^2}{c+a}-\frac{a^2}{c+a}\) \(=a-b+b-c+c-a=0\)
\(\Rightarrow\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}=\frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{c+a}\)
\(\Rightarrow2\left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\right)=\frac{a^2}{a+b}+\frac{b^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{b+c}+\frac{c^2}{c+a}+\frac{a^2}{c+a}\)\(\ge\frac{2ab}{a+b}+\frac{2bc}{b+c}+\frac{2ca}{c+a}\)
\(\Rightarrowđpcm\)
Dấu "=" \(\Leftrightarrow a=b=c\)
b) \(a^2b^2\left(a^2+b^2\right)=\frac{1}{2}\cdot ab\cdot2ab\cdot\left(a^2+b^2\right)\le\frac{1}{2}\cdot\frac{\left(a+b\right)^2}{4}\cdot\frac{\left(2ab+a^2+b^2\right)^2}{4}=2\)
Dấu "=" \(\Leftrightarrow a=b=1\)
1) \(xy\le\frac{\left(x+y\right)^2}{4}\)(cô si) ÁP DỤNG BẤT ĐẲNG THỨC TRÊN với a, b,c>0 TA CÓ
\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{\left(a+b\right)^2}{4\left(a+b\right)}+\frac{\left(b+c\right)^2}{4\left(b+c\right)}+\frac{\left(c+a\right)^2}{4\left(c+a\right)}.\)
\(=\frac{a+b}{4}+\frac{b+c}{4}+\frac{c+a}{4}=\frac{2\left(a+b+c\right)}{4}=\frac{a+b+c}{2}.\)
2) Với a,b,c >0 .XÉT \(\frac{a^2}{b}+b\ge2\sqrt{\frac{a^2}{b}.b}=2a\)(bất đẳng thức cô si)
\(\frac{b^2}{c}+c\ge2\sqrt{\frac{b^2}{c}.c}=2b\)
\(\frac{c^2}{a}+a\ge2\sqrt{\frac{c^2}{a}.a}=2c\)
\(\Rightarrow\frac{a^2}{b}+b+\frac{b^2}{c}+c+\frac{c^2}{a}+a\ge2a+2b+2c\)
\(\Leftrightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c\)
(đpcm)
\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{ab}{2\sqrt{ab}}+\frac{bc}{2\sqrt{bc}}+\frac{ca}{2\sqrt{ca}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}\le\frac{a+b+c}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)