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Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)
\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)
\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)
\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)
\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)
Dấu "=" xảy ra khi x=y=z
Bài 1
Cho a , b , c > 0 . CM : \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{a+b}{b+c}+\frac{b+c}{a+b}\left(1\right)\)
\(\Leftrightarrow\left(a+b\right)^2+\left(b+c\right)^2+\left(a+b\right)\left(b+c\right)\le\frac{a\left(a+b\right)\left(b+c\right)}{b}+\frac{b\left(a+b\right)\left(b+c\right)}{c}+\frac{c\left(a+b\right)\left(b+c\right)}{a}\)
\(=\frac{a^2c}{b}+a^2+ab+ac+\frac{b^2\left(a+b\right)}{c}+b^2+ab+c^2+bc+\frac{cb\left(b+c\right)}{a}\)
Mặt khác : \(\left(a+b\right)^2+\left(b+c\right)^2+\left(a+b\right)\left(b+c\right)=a^2+ac+c^2+3b^2+3ab+3bc\)
Do đó ta cần chứng minh :
\(\frac{a^2c}{b}+\frac{b^2\left(a+b\right)}{c}+\frac{cb\left(b+c\right)}{a}\ge2b^2+2bc+ab\left(2\right)\)
\(VT=\frac{a^2c}{b}+\frac{b^2\left(a+b\right)}{c}+\frac{cb\left(b+c\right)}{a}=\frac{1}{2}\left(\frac{a^2c}{b}+\frac{b^3}{c}\right)+\frac{1}{2}\left(\frac{a^2c}{b}+\frac{c^2b}{a}\right)+\frac{1}{2}\left(\frac{b^3}{c}+\frac{c^2b}{a}\right)+b^2\left(\frac{c}{a}+\frac{a}{c}\right)\)
\(\ge ab+\sqrt{ac^3}+\sqrt{\frac{b^4c}{a}}+2b^2\ge ab+2bc+2b^2=VP\)
Dấu " = " xảy ra khi a=b=c
Bài 2 :
Vì x , y , z > 0 ta có :
Áp dụng BĐT Cô - si đối với 2 số dương \(\frac{x^2}{y+z}\) và \(\frac{y+z}{4}\)
ta được :
\(\frac{x^2}{y+z}+\frac{y+z}{4}\ge2\sqrt{\frac{x^2}{y+z}.\frac{y+z}{4}}=2.\frac{x}{2}=x\left(1\right)\) .
Tương tự ta cũng có :
\(\frac{y^2}{x+z}+\frac{x+z}{4}\ge y\left(2\right);\frac{z^2}{x+y}+\frac{x+y}{4}\ge z\left(3\right)\)
Cộng theo vế (1) , (2) và (3) ta được :
\(\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\frac{x+y+z}{2}\ge x+y+z\Rightarrow P\ge\left(x+xy+z\right)-\frac{x+y+z}{2}=1\)
Dấu " = " xảy ra \(\Leftrightarrow x=y=z=\frac{2}{3}\)
Vậy \(P=1\Leftrightarrow x=y=z=\frac{2}{3}\)
Áp dụng Cauchy Schwarz dạng Engel ta có :
\(\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{a+b+c}{5}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Cho \(a=b=c\)
\(\Rightarrow2\left(\frac{a}{a+2a}+\frac{a}{a+2a}+\frac{a}{a+2a}\right)\ge1+\frac{a}{a+2a}+\frac{a}{a+2a}+\frac{a}{a+2a}\)
\(\Leftrightarrow2\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)\ge1+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\)
\(\Leftrightarrow2\ge2\) ( Đúng)
\(\Rightarrow2\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\ge1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
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 \(a=b=c=1\)
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