cho a+b+c=3 chứng minh 1/(2+a^2b)+1/(2+b^2c)+1(/2+c^2a) >= 1
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\(\dfrac{a^2b^2}{2a^2+b^2+3a^2b^2}=\dfrac{a^2b^2}{\left(a^2+b^2\right)+\left(a^2+a^2b^2\right)+2a^2b^2}\le\dfrac{a^2b^2}{2ab+2a^2b+2a^2b^2}=\dfrac{ab}{2\left(1+a+ab\right)}\)
Tương tự và cộng lại;
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{bc}{1+b+bc}+\dfrac{ca}{1+c+ca}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{abc}{a+ab+abc}+\dfrac{ab.ca}{ab+abc+ab.ca}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{1}{a+ab+1}+\dfrac{a}{ab+1+a}\right)=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Bài 1:
\(BDT\Leftrightarrow\sqrt{\frac{3}{a+2b}}+\sqrt{\frac{3}{b+2c}}+\sqrt{\frac{3}{c+2a}}\le\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)
\(\Leftrightarrow\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\ge\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)
Áp dụng BĐT Cauchy-Schwarz và BĐT AM-GM ta có:
\(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{b}}\ge\frac{9}{\sqrt{a}+\sqrt{2}\cdot\sqrt{2b}}\ge\frac{9}{\sqrt{\left(1+2\right)\left(a+2b\right)}}=\frac{3\sqrt{3}}{\sqrt{a+2b}}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{c}}\ge\frac{3\sqrt{3}}{\sqrt{b+2c}};\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{a}}\ge\frac{3\sqrt{3}}{\sqrt{c+2a}}\)
Cộng theo vế 3 BĐT trên ta có:
\(3\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\ge3\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)
\(\Leftrightarrow\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\ge\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)
Đẳng thức xảy ra khi \(a=b=c\)
Bài 2: làm mãi ko ra hình như đề sai, thử a=1/2;b=4;c=1/2
Bài 2/
\(\frac{bc}{a^2b+a^2c}+\frac{ca}{b^2c+b^2a}+\frac{ab}{c^2a+c^2b}\)
\(=\frac{b^2c^2}{a^2b^2c+a^2c^2b}+\frac{c^2a^2}{b^2c^2a+b^2a^2c}+\frac{a^2b^2}{c^2a^2b+c^2b^2a}\)
\(=\frac{b^2c^2}{ab+ac}+\frac{c^2a^2}{bc+ba}+\frac{a^2b^2}{ca+cb}\)
\(\ge\frac{\left(bc+ca+ab\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)
\(\ge\frac{3\sqrt[3]{ab.bc.ca}}{2}=\frac{3}{2}\)
Dấu = xảy ra khi \(a=b=c=1\)
Lời giải:
Ta có: \(a^2b+b^2c+c^2a\geq \frac{9a^2b^2c^2}{1+2a^2b^2c^2}\)
\(\Leftrightarrow (a^2b+b^2c+c^2a)(1+2a^2b^2c^2)\geq 9a^2b^2c^2\)
\(\Leftrightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^3+2a^3b^2c^4\geq 3a^2b^2c^2(a+b+c)(*)\)
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Áp dụng BĐT AM-GM ta có:
\(a^2b+a^4b^3c^2+a^3b^2c^4\geq 3\sqrt[3]{a^9b^6c^6}=3a^3b^2c^2\)
\(b^2c+a^2b^4c^3+a^4b^3c^2\geq 3a^2b^3c^2\)
\(c^2a+a^3b^2c^4+a^2b^4c^3\geq 3a^2b^2c^3\)
Cộng theo vế:
\(\Rightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^3+2a^3b^2c^4\geq 3a^2b^2c^2(a+b+c)\)
Vậy $(*)$ đúng
Do đó ta có đpcm
Dấu bằng xảy ra khi $a=b=c=1$
Ta có :
\(a^2b+b^2c+c^2a\ge\frac{9a^2b^2c^2}{1+2a^2b^2c^2}\)
\(\Leftrightarrow\left(a^2b+b^2c+c^2a\right)\left(1+2a^2b^2c^2\right)\ge9a^2b^2c^2\)
\(\Leftrightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^{3v}+2a^3b^2c^4\ge3a^2b^2c^2\left(a+b+c\right)\)(*)
Áp dụng BĐT AM-GM ta có:
\(a^2b+a^4b^3c^2+a^3b^2c^4\ge3\sqrt[3]{a^9b^6c^6}=3a^3b^2c^2\)
\(b^2c+a^2b^4c^3+a^4b^3c^2\ge3a^2b^3c^2\)
\(c^2a+a^3b^2c^4+a^2b^4c^4\ge3a^2b^2c^3\)
Cộng theo vế
\(\Rightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^3+2a^3b^2c^4\ge3a^2b^2c^2\left(a+b+c\right)\)
Vậy $(*)$ đúng
Do đó ta có đpcm
#Cừu
VT=2a2b2+2a2c2+2b2c2-a4-b4-c4
=a2b2+a2c2+b2c2+a2.(b2-a2)+b2.(c2-b2)+c2.(a2-c2)
=a2b2+a2c2+b2c2+a2.(b+a)(b-a)+b2.(c+b)(c-b)+c2.(a+c)(a-c)
Ta lại có : a+b>c=>a-c>-b
b+c>a=>b-a>-c
c+a>b=>c-b>-a
(BĐT tam giác)
=>VT>a2b2+a2c2+b2c2+a2.c.(-c)+b2.a.(-a)+c2.b.(-b)
=0
=>VT>0 =>dpcm