Xin các mems giúp bài toán này với:
Cho a, b, c > 0 và a2 + b2 + c2 = 1. CMR: \(\sqrt{1-ab}+\sqrt{1-bc}+\sqrt{1-ca}\ge\sqrt{6}\)
Cho a, b, c là các số thực dương thỏa mãn 2(a2 +b2 +c2) = a+b+c+3. Chứng minh rằng:
\(\dfrac{1}{\sqrt{a^4+a^2+1}}\)+ \(\dfrac{1}{\sqrt{b^4+b^2+1}}\)+ \(\dfrac{1}{\sqrt{c^4+c^2+1}}\) \(\ge\sqrt{3}\)
mng giúp mình nhé, cảm ơnn
Cho a,b,c>0 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
CMR \(\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{abc}\)
cho a,b,c \(\ge\)0 thỏa a+b+c=1.CMR
\(\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge1+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
Ta chứng minh: \(\sqrt{a+bc}\ge a+\sqrt{bc}\)
Thật vậy, ta có:
\(a+bc\ge a^2+2a\sqrt{bc}+bc\)
\(\Leftrightarrow a\ge a^2+2a\sqrt{bc}\)
\(\Leftrightarrow1\ge a+2\sqrt{bc}\)
\(\Leftrightarrow a+b+c\ge a+2\sqrt{bc}\)
\(\Leftrightarrow b+c\ge2\sqrt{bc}\)(Đúng theo Cauchy)
Tương tự: \(\sqrt{b+ca}\ge b+\sqrt{ca}\)
\(\sqrt{c+ab}\ge c+\sqrt{ab}\)
Cộng vế theo vế các BĐT vừa chứng minh ta được đpcm.
Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Giả sử a;b;c là dộ dài 3 cạnh của 1 tam giác. CMR :
\(\frac{1}{\sqrt{ab+ca}}+\frac{1}{\sqrt{bc+ab}}+\frac{1}{\sqrt{ca+bc}}\ge\frac{1}{\sqrt{a^2+bc}}+\frac{1}{\sqrt{b^2+ca}}+\frac{1}{\sqrt{c^2+ab}}\)
Cho a,b,c>0 thỏa mãn\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=1\). CMR
\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{a+b}\ge\dfrac{1}{2}\)
Áp dụng BĐT BSC:
\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}\)
\(=\dfrac{a+b+c}{2}\)
\(\ge\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\dfrac{1}{2}\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{3}\)
CMR: với a, b, c > 0 thì:
\(\sqrt{\frac{a}{bc}}+\sqrt{\frac{b}{ca}}+\sqrt{\frac{c}{ab}}\ge\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}}\)
Áp dụng bđt Cauchy, ta có:
\(\sqrt{\frac{a}{bc}}\)+\(\sqrt{\frac{b}{ca}}\)≥ \(2\sqrt{\sqrt{\frac{ab}{abc^2}}}\)= \(2\sqrt{\sqrt{\frac{1}{c^2}}}\)= \(2\sqrt{\frac{1}{c}}\) (vì c>0)
Tương tự: \(\sqrt{\frac{b}{ca}}\)+\(\sqrt{\frac{c}{ab}}\)≥ \(2\sqrt{\frac{1}{a}}\)
\(\sqrt{\frac{c}{ab}}\)+\(\sqrt{\frac{a}{bc}}\)≥ \(2\sqrt{\frac{1}{b}}\)
Cộng vế theo vế của các bđt với nhau, ta có: \(2\)\(\left(\sqrt{\frac{a}{bc}}+\sqrt{\frac{b}{ca}}+\sqrt{\frac{c}{ab}}\right)\text{≥}\)\(2\left(\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}}\right)\)
<=> \(\sqrt{\frac{a}{bc}}+\sqrt{\frac{b}{ca}}+\sqrt{\frac{c}{ab}}\text{≥}\)\(\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}}\)(đpcm)
Dấu "=" xảy ra <=> a = b = c
Bài 1 :Cho a,b,c dương thỏa mãn a+b+c=2
CMR \(\frac{bc}{\sqrt{3a^2+4}}+\frac{ca}{\sqrt{3b^2+4}}+\frac{ab}{\sqrt{3c^2+4}}\ge\frac{\sqrt{3}}{3}\)
Bài 2:Cho a,b,c>0. CMR
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm
cho các số thực dương a,b,c thỏa mãn abc=1. CMR:\(\frac{1}{\sqrt{a^5+b^2+ab+6}}+\frac{1}{\sqrt{b^5+c^2+bc+6}}+\frac{1}{\sqrt{c^5+a2+ca+6}}\)\(\le\)\(1\)
Ta có:\(a^5+ab+b^2\ge3a^2b\)
Tương tự ta có:
\(VT\le\frac{1}{\sqrt{3ab\left(a+2c\right)}}+\frac{1}{\sqrt{3bc\left(b+2a\right)}}+\frac{1}{\sqrt{3ca\left(c+2b\right)}}\)
\(=\frac{1}{\sqrt{3}}\left(\sqrt{\frac{c}{c+2a}}+\sqrt{\frac{a}{b+2a}}+\sqrt{\frac{b}{2b+c}}\right)\)
Ta cũng có:\(a+2c=a+c+c\ge\frac{1}{3}\left(\sqrt{a}+2\sqrt{c}\right)^2\)
\(\Rightarrow VT\le\frac{\sqrt{c}}{\sqrt{a}+2\sqrt{c}}+\frac{\sqrt{a}}{\sqrt{b}+2\sqrt{a}}+\frac{\sqrt{b}}{\sqrt{c}+2\sqrt{b}}\)
Đặt \(x=\frac{\sqrt{a}}{\sqrt{c}};y=\frac{\sqrt{b}}{\sqrt{a}};z=\frac{\sqrt{c}}{\sqrt{b}};xyz=1\)
\(\Rightarrow VT\le\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\)
Giả sử \(xy\le1\) thì \(z\ge1\)
Ta có: \(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\frac{1}{2}\left(\frac{1}{\frac{x}{2}+1}+\frac{1}{\frac{y}{2}+1}\right)+\frac{1}{z+2}\)
\(\le\frac{1}{1\frac{\sqrt{xy}}{2}}+\frac{1}{z+2}\le1\)(Đpcm)
Dấu = khi \(a=b=c=1\)
sao chứng minh đc \(a^5+ab+b^2\ge3a^2b\)vậy bạn
a,b dương thì áp dụng Cô si
a5+ab+b2 \(\ge3\sqrt[3]{a^5\cdot a\cdot b\cdot b^2}=3\sqrt[3]{a^6b^3}=3a^2b\)
hiểu chưa ???
Cho các số thực dương a,b,c thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\). CMR:
\(\frac{a+b}{\sqrt{ab+c}}+\frac{b+c}{\sqrt{bc+a}}+\frac{c+a}{\sqrt{ca+b}}\ge3\sqrt[6]{abc}\)
Giải:
\(GT\Leftrightarrow ab+bc+ca\ge abc\)
\(\Rightarrow ab\le\frac{ab+bc+ca}{c}\)
\(\Rightarrow\frac{a+b}{\sqrt{ab+c}}\ge\frac{a+b}{\sqrt{\frac{ab+bc+ca}{c}+c}}=\frac{\left(a+b\right)\sqrt{c}}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
Tương tự rồi cộng lại: \(VT\ge\frac{\left(a+b\right)\sqrt{c}}{\sqrt{\left(c+a\right)\left(c+b\right)}}+\frac{\left(b+c\right)\sqrt{a}}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{\left(c+a\right)\sqrt{c}}{\sqrt{\left(b+a\right)\left(b+c\right)}}\)\(\ge3\sqrt[3]{\sqrt{abc}}=3\sqrt[6]{abc}\)
Lần sau mấy bạn hỏi bài thì đăng lên nhé!
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