\(P=\dfrac{a}{a^2+bc}+\dfrac{b}{b^2+ca}+\dfrac{c}{c^2+ab}\)
\(\le\dfrac{a}{2a\sqrt{bc}}+\dfrac{b}{2b\sqrt{ca}}+\dfrac{c}{2c\sqrt{ab}}\)
\(=\dfrac{a\sqrt{bc}}{2abc}+\dfrac{b\sqrt{ca}}{2abc}+\dfrac{c\sqrt{ab}}{2abc}\)
\(\le\dfrac{2a^2+b^2+c^2}{8abc}+\dfrac{2b^2+a^2+c^2}{8abc}+\dfrac{2c^2+b^2+a^2}{8abc}\)
\(=\dfrac{4\left(a^2+b^2+c^2\right)}{8abc}=\dfrac{1}{2}\)
Cho a + b + c = 1 (a,b,c khác 1,2). Chứng minh
\(\dfrac{c+ab}{a^2+b^2+abc-1}+\dfrac{a+bc}{b^2+c^2+abc-1}+\dfrac{b+ac}{a^2+c^2+abc-1}=\dfrac{bc+ac+ab+8}{\left(a-2\right)\left(b-2\right)\left(a-2\right)}\)
giải phương trình nghiệm nguyên:
\(\left\{{}\begin{matrix}ab+bc+ca=0\\\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+\dfrac{3}{4}=0\end{matrix}\right.\)
Xét:
\(\dfrac{c}{a-b}.\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{c}{a-b}\left(\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{c}{a-b}.\dfrac{b^2-bc+ac-a^2}{ab}=1+\dfrac{c}{a-b}.\dfrac{c\left(a-b\right)-\left(a^2-b^2\right)}{ab}=1+\dfrac{c}{a-b}.\dfrac{\left(c-a-b\right)\left(a-b\right)}{ab}=1+\dfrac{c^2-c\left(a+b\right)}{ab}=1+\dfrac{2c^2}{ab}=1+\dfrac{2c^3}{abc}\)
CMTT cộng theo vế:
\(BTCCM=3+\dfrac{2\left(a^3+b^3+c^3\right)}{abc}=\dfrac{6\left(a^3+b^3+c^3\right)}{3abc}\)
Mà Khi \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\) ( tự cm,ez)
Vậy \(BTCCM=3+6=9\left(đpcm\right)\)
CMR nếu \(\left(a^2-bc\right).\left(b-abc\right)=\left(b^2-ac\right).\left(a-abc\right)\) và các số a, b, c, a-b khác 0 thì \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c\)
Chứng minh đẳng thức:
\(\dfrac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\dfrac{1}{\left(c-a\right)\left(b^2+ba-c^2-ca\right)}+\dfrac{1}{\left(a-b\right)\left(c^2+cb-a^2-ab\right)}=0\)
Tổng \(S=\dfrac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^2-ac}{\left(b+c\right)\left(b+a\right)}+\dfrac{c^2-ab}{\left(c+a\right)\left(c+b\right)}=\)
Cho \(\left(a^2-bc\right)\left(b-abc\right)=\left(b^2-ac\right)\left(a-abc\right);abc\ne0;a\ne b\)
CMR:\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c\)
Rút gọn biểu thức: \(B=\left(ab+bc+ca\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-abc.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
Hứa tặng GP nha :))
I. BĐT:
1.Cho a,b,c là độ dài của ba cạnh tam giác CMR:
\(\left(a\right)a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)
\(\left(b\right)\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\ge3\)
\(\left(c\right)\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\)
2. Cho a, b, c, d > 0 và abcd = 1 CMR: \(a^2+b^2+c^2+d^2+ab+cd\ge6\)
3. \(\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-6\right)+9\ge0\)
4. \(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\le\dfrac{a+b +c}{2}\)
