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Ta có: \(P=1+\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)+\left(\frac{1}{a^3b^3}+\frac{1}{b^3c^3}+\frac{1}{a^3c^3}+\frac{1}{a^3b^3c^3}\right)\)
\(P\ge a+\frac{3}{abc}+\frac{3}{a^2b^2c^2}+\frac{1}{a^3b^3c^3}=\left(1+\frac{1}{abc}\right)^3\) (BĐT Cosi cho 3 số dương)
Theo BĐT Cosi \(abc\le\left(\frac{a+b+c}{3}\right)^3=8̸\)\(\Rightarrow abc\le8\Rightarrow\frac{1}{abc}\ge\frac{1}{8}\)
Vậy \(P\ge\left(1+\frac{1}{8}\right)^3=\frac{729}{512}\)
Dấu "=" xảy ra khi a=b=c=2
Cho a,b,c > 0 thỏa mãn a+b+c=1. Tìm Min \(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}+\frac{1}{9abc}\)
\(A\ge\frac{9}{a+2+b+2+c+2}+\frac{1}{9abc}\)
\(\Rightarrow A\ge\frac{9}{7}+\frac{1}{9abc}\)
Theo BĐT AM-GM ta có: \(1=a+b+c\ge3\sqrt[3]{abc}\)
\(\Rightarrow abc\le\frac{1}{27}\)
\(\Rightarrow\frac{1}{9abc}\ge3\)
Do đó ta có:
\(A\ge\frac{9}{7}+3=\frac{30}{7}\)
a, Áp dụng \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)
Áp dụng \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\forall x,y>0\)
Ta có: \(A=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2\ge\frac{\left(2+\frac{1}{a}+\frac{1}{b}\right)^2}{2}\ge\frac{\left(2+\frac{4}{a+b}\right)^2}{2}\ge\frac{\left(2+4\right)^2}{2}=18\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
b, Áp dụng \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)
Áp dụng \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\forall x,y,z>0\)
Ta có: \(B=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2+\left(1+\frac{1}{c}\right)^2\ge\frac{\left(3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\ge\frac{\left(3+\frac{9}{a+b+c}\right)^2}{3}\ge\frac{\left(3+6\right)^2}{3}=27\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{2}\)
* Các BĐT phụ bạn tự CM nha! Chúc bạn học tốt
a.
\(A=\frac{1}{ab}+\frac{1}{a^2+b^2}=\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\frac{1}{2ab}\)
\(\ge\frac{4}{a^2+2ab+b^2}+\frac{1}{2ab}\ge\frac{4}{\left(a+b\right)^2}+\frac{1}{\frac{\left(a+b\right)^2}{2}}=6\)
Dấu "=" khi \(a=b=\frac{1}{2}\)
b.
\(B=\frac{2}{ab}+\frac{3}{a^2+b^2}=3\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\frac{1}{2ab}\)
\(\ge3\cdot\frac{4}{\left(a+b\right)^2}+\frac{1}{\frac{\left(a+b\right)^2}{2}}=14\)
Dấu "=" khi \(a=b=\frac{1}{2}\)
c.
Ta có:
\(x^2+y^2\ge2xy\)
\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)
\(\Leftrightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\) với mọi x,y
Áp dụng ta có:
\(C=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\ge\frac{\left(a+b+\frac{1}{a}+\frac{1}{b}\right)^2}{2}\ge\frac{\left(1+\frac{4}{a+b}\right)^2}{2}=\frac{25}{2}\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
2.
Áp dụng bất đẳng thức Bunhiacopxki ta có:
\(\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2\right]\left[\left(\frac{a}{\sqrt{x}}\right)^2+\left(\frac{b}{\sqrt{y}}\right)^2\right]\ge\left(\sqrt{x}\cdot\frac{a}{\sqrt{x}}+\sqrt{y}\cdot\frac{b}{\sqrt{y}}\right)^2\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{a^2}{x}+\frac{b^2}{y}\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)
Áp dụng nó ta chứng minh được:
\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b\right)^2}{x+y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
Áp dụng vào bài làm:
\(D=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{ab+ca}+\frac{b^2}{bc+ab}+\frac{c^2}{ca+bc}\)
\(\ge\frac{\left(a+b+c\right)^2}{ab+ca+bc+ab+ca+bc}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
Bài làm:
Bài 1:
Ta có: \(T=8x^2-4x+\frac{1}{4x^2}+15\)
\(=\left(4x^2-4x+1\right)+\left(4x^2+\frac{1}{4x^2}\right)+14\)
\(=\left(2x-1\right)^2+\left(4x^2+\frac{1}{4x^2}\right)+14\)\(\ge0+2\sqrt{4x^2.\frac{1}{4x^2}}+14=2+14=16\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(2x-1\right)^2=0\\4x^2=\frac{1}{4x^2}\end{cases}\Rightarrow x=\frac{1}{2}}\)
Vậy \(Min\left(T\right)=16\)khi \(x=\frac{1}{2}\)
Bài 2:
Ta có: \(ab+bc+ca=3abc\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=3\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\left(1\right)\)
Ta xét \(\frac{a^2}{c\left(c^2+a^2\right)}=\frac{\left(c^2+a^2\right)-c^2}{c\left(c^2+a^2\right)}=\frac{1}{c}-\frac{c}{c^2+a^2}=\frac{1}{c}-\frac{1}{a}.\frac{ac}{c^2+a^2}\ge\frac{1}{c}-\frac{1}{a}.\frac{ac}{2ac}=\frac{1}{c}-\frac{1}{2}a\)
Tương tự ta chứng minh được: \(\frac{b^2}{a\left(a^2+b^2\right)}\ge\frac{1}{a}-\frac{1}{2}b\)và \(\frac{c^2}{b\left(b^2+c^2\right)}\ge\frac{1}{b}-\frac{1}{2}c\)
Cộng vế 3 bất đẳng thức trên lại ta được:
\(P\ge\frac{1}{c}-\frac{1}{2}a+\frac{1}{a}-\frac{1}{2}b+\frac{1}{b}-\frac{1}{2}c\)\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{2}.3=\frac{3}{2}\left(theo\left(1\right)\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}a^2=b^2\\b^2=c^2\\c^2=a^2\end{cases}\Rightarrow a=b=c=1}\)
Vậy \(Min\left(P\right)=\frac{3}{2}\)khi \(a=b=c=1\)
Học tốt!!!!
Theo AM-GM có :
\(9a+\frac{1}{a}\ge2\sqrt{9a.\frac{1}{a}}=6\)
\(9b+\frac{1}{b}\ge2\sqrt{9b.\frac{1}{b}}=6\)
\(9c+\frac{1}{c}\ge2\sqrt{9c.\frac{1}{c}}=6\)
Cộng vế theo vế :
\(9\left(a+b+c\right)+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge18\)
\(8\left(a+b+c\right)+\left(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\right)\ge18\)
\(\left(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\right)\ge10\)
Vậy \(Min=10\)
\(\Leftrightarrow a=b=c=\frac{1}{3}\)
S = a+b+c + (1/a + 1/b + 1/c)
>= (a+b+c) + 9/a+b+c
= [ (a+b+c) + 9/4.(a+b+c) ] + 27/4.(a+b+c)
>= \(2\sqrt{\left(a+b+c\right).\frac{9}{4.\left(a+b+c\right)}}\) + 27/(4.3/2)
= 3 + 9/2
= 15/2
Dấu "=" xảy ra <=> a=b=c=1/2
Vậy ......
Tk mk nha
Do a ; b ; c > 0 , áp dụng BĐT Cô - si cho 3 số , ta có :
\(a+b+c\ge3\sqrt[3]{abc};\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
\(\Rightarrow3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)