cho M =\(\frac{b-c}{a^2-ac-ab+bc}+\frac{c-a}{b^2-ab-cb+ca}+\frac{a-b}{c^2-bc-ac+ab}\) và N=\(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\) cmr M=2N
cho a,b,c > 0 thỏa mãn \(a^2+b^2+c^2=1\) . Cmr:
\(\sqrt{\frac{ab+2c^2}{1+ab-c^2}}+\sqrt{\frac{bc+2a^2}{1+bc-a^2}}+\sqrt{\frac{ca+2b^2}{1+ac-b^2}}\ge2+ab+bc+ca\)
\(\sqrt{\frac{ab+2c^2}{1+ab-c^2}}=\sqrt{\frac{ab+2c^2}{a^2+b^2+ab}}=\frac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(a^2+b^2+ab\right)}}\ge\frac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\ge\frac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)
Tương tự: \(\sqrt{\frac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\) ; \(\sqrt{\frac{ca+2b^2}{1+ac-b^2}}\ge ca+2b^2\)
Cộng vế với vế:
\(VT\ge2\left(a^2+b^2+c^2\right)+ab+bc+ca=2+ab+bc+ca\)
CMR: \(\frac{a^2+b^2+c^2}{ab+bc+ac} + \frac{1}{3} \geq \frac{8}{9}(\frac{a}{b+c} + \frac{b}{a+c} +\frac{c}{a+b})\)
CMR:\((1+a+b+c)(1+ab+bc+ac) \geq 4\sqrt{2(a+bc)(b+ac)(c+ab)}\)
cho a,b,c > 0 thỏa mãn ab+bc+ca=1. Cmr:
\(a+b+c+\frac{ab}{b+c}+\frac{bc}{c+a}+\frac{ca}{a+b}\ge\frac{3\sqrt{3}}{2}\)
Lời giải:
Ta thấy:
\(\text{VT}=(a+\frac{ca}{a+b})+(b+\frac{ab}{b+c})+(c+\frac{bc}{c+a})\)
\(=\frac{a(a+b+c)}{a+b}+\frac{b(a+b+c)}{b+c}+\frac{c(a+b+c)}{c+a}\)
\(=(a+b+c)\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\)
\(\geq (a+b+c).\frac{(a+b+c)^2}{a^2+ab+b^2+bc+c^2+ac}=\frac{(a+b+c)^3}{a^2+b^2+c^2+ab+bc+ac}\) (theo BĐT Cauchy-Schwarz)
Có:
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=a^2+b^2+c^2+2$
$\Rightarrow a+b+c=\sqrt{a^2+b^2+c^2+2}=\sqrt{t+2}$ với $t=a^2+b^2+c^2$
Do đó:
$\text{VT}\geq \frac{\sqrt{(t+2)^3}}{t+1}$ \(=\sqrt{\frac{(t+2)^3}{(t+1)^2}}\)
Áp dụng BĐT AM-GM:
\((t+2)^3=\left(\frac{t+1}{2}+\frac{t+1}{2}+1\right)^3\geq 27.\frac{(t+1)^2}{4}\)
\(\Rightarrow \text{VT}=\sqrt{\frac{(t+2)^3}{(t+1)^2}}\geq \sqrt{\frac{27}{4}}=\frac{3\sqrt{3}}{2}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{\sqrt{3}}$
Quay lại diễn đàn trong thinh lặng:))
Chứng minh: $$\left( a+{\frac {ab}{b+c}}+b+{\frac {bc}{c+a}}+c+{\frac {ca}{a+b}}
\right) ^{2}-{\frac {27\,ab}{4}}-{\frac {27\,ca}{4}} \geqq {\frac {27\,bc}{
4}}$$
Sau khi quy đồng, cần chứng minh$:$
$$\frac{1}{2} \sum\limits_{cyc} \left( 5\,{a}^{4}{b}^{2}+8\,{a}^{3}{b}^{3}+7\,{a}^{2}{b}^{4}+98\,{a}^
{2}{b}^{3}c+99\,{a}^{2}{b}^{2}{c}^{2}+124\,{a}^{2}b{c}^{3}+34\,a{b}^{4
}c+130\,a{b}^{3}{c}^{2}+26\,{b}^{4}{c}^{2}+44\,{b}^{3}{c}^{3}+{c}^{6}
\right) \left( a-b \right) ^{2} \geqq 0$$
Cho các số dương a, b, c thỏa mãn ab+bc+ca=1.
CMR: \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge3+\sqrt{\frac{\left(a+b\right)\left(a+c\right)}{a^2}}+\sqrt{\frac{\left(b+c\right)\left(b+a\right)}{b^2}}+\sqrt{\frac{\left(c+a\right)\left(c+b\right)}{c^2}}\)
\(VT=\frac{ab+bc+ca}{ab}+\frac{ab+bc+ca}{bc}+\frac{ab+bc+ca}{ca}\)
\(=3+\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}+\frac{b\left(c+a\right)}{ca}\)(1)
Theo BĐT AM-GM: \(\frac{1}{2}\left[\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}\right]\ge\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{b^2}}\)
Tương tự: \(\frac{1}{2}\left[\frac{a\left(b+c\right)}{bc}+\frac{b\left(c+a\right)}{ca}\right]\ge\sqrt{\frac{\left(a+c\right)\left(b+c\right)}{c^2}}\)
\(\frac{1}{2}\left[\frac{c\left(a+b\right)}{ab}+\frac{b\left(c+a\right)}{ca}\right]\ge\sqrt{\frac{\left(a+c\right)\left(a+b\right)}{a^2}}\)
Cộng theo vế 3 BĐT trên rồi thay vào 1 ta sẽ thu được đpcm.
Cho a, b, c > 0 thỏa mãn a+b+c=1
Tính \(P=\left(\frac{a-bc}{a+bc}+\frac{b-ac}{b+ac}+\frac{c-ab}{c+ab}\right):\frac{ab+bc+ca+3abc}{ab+bc-abc}.\)
\(\frac{a^2b+bc^2-1}{ac\left(a+c\right)}+\frac{b^2c+ca^2-1}{ab\left(a+b\right)}+\frac{c^2a+ab^2-1}{bc\left(b+c\right)}\)
\(=\frac{a^2b^2+b^2c^2-b}{a+c}+\frac{b^2c^2+c^2a^2-c}{a+b}+\frac{c^2a^2+a^2b^2-a}{b+c}\)
\(=\frac{\frac{1}{a^2}-\frac{1}{ac}+\frac{1}{c^2}}{a+c}+\frac{\frac{1}{b^2}-\frac{1}{ab}+\frac{1}{a^2}}{a+b}+\frac{\frac{1}{c^2}-\frac{1}{bc}+\frac{1}{b^2}}{b+c}\ge\frac{1}{ac\left(a+c\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ab\left(b+a\right)}\)
\(=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
Cho 3 số dương a,b,c thỏa mãn a+b+c = 1/2 và a^2+b^2+c^2+ab+bc+ca =1/6. tính giá trị BT : P = \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
Cho a,b,c thỏa mãn \(\frac{a^3}{a^{^2}+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}=1006\).Tính giá trị của biểu thức \(M=\frac{a^3+b^3}{a^2+ab+b^2}+\frac{b^3+c^3}{b^2+bc+c^2}+\frac{c^3+a^3}{c^2+ca+a^2}\)
Cho \(S=1.2.3+2.3.4+3.4.5+...+n\left(n+1\right)\left(n+2\right)\). CMR \(4S+1\)là số chính phương
Cho a,b,c>0 thỏa mãn ab+bc+ac=1. CMR \(\frac{a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\le\frac{3}{2}\)
\(VT=\frac{a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)
\(=\frac{a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ca}}+\frac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(=\frac{a}{\sqrt{\left(a+b\right)\left(c+a\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(b+c\right)\left(c+a\right)}}\)
\(=\sqrt{\frac{a}{a+b}.\frac{a}{c+a}}+\sqrt{\frac{b}{a+b}.\frac{b}{b+c}}+\sqrt{\frac{c}{b+c}.\frac{c}{c+a}}\)
\(\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{c+a}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{c}{c+a}\right)\)
\(=\frac{1}{2}.3=\frac{3}{2}\)
Dấu = xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
cho 3 số a,b,c thỏa mãn điều kiện \(\frac{1}{bc-a^2}+\frac{1}{ca-b^2}+\frac{1}{ab-c^2}=0\)
CMR: \(\frac{a}{\left(bc-a^2\right)^2}+\frac{b}{\left(ca-b^2\right)^2}+\frac{c}{\left(ab-c^2\right)^2}=0\)
từ giả thiết ta có
\(\frac{1}{bc-a^2}=\frac{1}{b^2-ca}+\frac{1}{c^2-ab}=\frac{c^2-ab+b^2-ca}{\left(b^2-ca\right)\left(c^2-ab\right)}\)
Nhân hai vế với \(\frac{a}{bc-a^2}\) ta có:
\(\frac{a}{\left(bc-a^2\right)^2}=\frac{ac^2-a^2b+ab^2-ca^2}{\left(bc-a^2\right)\left(b^2-ca\right)\left(c^2-ab\right)}\)
làm tương tự với hai số hạng còn lại ta được:
\(\frac{b}{\left(ca-b^2\right)^2}=\frac{bc^2-ab^2+a^2b-b^2c}{\left(bc-a^2\right)\left(b^2-ca\right)\left(c^2-ab\right)}\);\(\frac{c}{\left(ab-c^2\right)^2}=\frac{b^2c-c^2a+a^2c-bc^2}{\left(bc-a^2\right)\left(b^2-ca\right)\left(c^2-ab\right)}\)
cộng ba vế của đẳng thức trên ta được kq là 0
cách kia dài quá
Đặt \(x=bc-a^2;y=ac-b^2;z=ab-c^2\)
Suy ra cần chứng minh \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\) thì \(\frac{a}{x^2}+\frac{b}{y^2}+\frac{c}{z^2}=0\)
Xét \(T=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)\).....