cho a,b,c>0 và a+b+c=1. tìm min F= \(\frac{a}{1+b-a}\)+\(\frac{b}{1+c-b}\)+\(\frac{c}{1+a-c}\)
cho a,b,c >0 và a+b+c=1 tìm min F =\(\frac{a^6}{b^3+c^3}\)+ \(\frac{b^6}{c^3+a^3}\)+\(\frac{c^6}{a^3+b^3}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(F=\frac{a^6}{b^3+c^3}+\frac{b^6}{c^3+a^3}+\frac{c^6}{a^3+b^3}\)
\(\ge\frac{\left(a^3+b^3+c^3\right)^2}{2\left(a^3+b^3+c^3\right)}=\frac{a^3+b^3+c^3}{2}\)
Áp dụng BĐT AM-GM ta có:
\(a^3+\frac{1}{27}+\frac{1}{27}\ge3\sqrt[3]{a^3\cdot\frac{1}{27}\cdot\frac{1}{27}}=3\cdot\frac{a}{9}=\frac{a}{3}\)
Tương tự ta cũng có: \(b^3+\frac{1}{27}+\frac{1}{27}\ge\frac{b}{3};c^3+\frac{1}{27}+\frac{1}{27}\ge\frac{c}{3}\)
\(\Rightarrow a^3+b^3+c^3+\frac{2}{9}\ge\frac{a+b+c}{3}=\frac{1}{3}\Rightarrow a^3+b^3+c^3\ge\frac{1}{9}\)
\(\Rightarrow F\ge\frac{a^3+b^3+c^3}{2}\ge\frac{\frac{1}{9}}{2}=\frac{1}{18}\)
Xảy ra khi \(a=b=c=\frac{1}{3}\)
Cho a,b,c >0 và a+b+c=3
Tìm min \(P=\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\)
\(P=a-\frac{ab^2}{1+b^2}+b-\frac{bc^2}{1+c^2}+c-\frac{ca^2}{1+a^2}\)
\(\ge a-\frac{ab^2}{2b}+b-\frac{bc^2}{2c}+c-\frac{ca^2}{2c}\) (AM-GM)
\(\ge a-\frac{ab}{2}+b-\frac{bc}{2}+c-\frac{ac}{2}\ge\left(a+b+c\right)-\frac{\left(a+b+c\right)^2}{6}\ge3-\frac{3}{2}=\frac{3}{2}\)
Vay MinP=3/2 dau = xay ra khi a=b=c=1
\(B=\frac{a+bc}{b+c}+\frac{b+ac}{a+c}+\frac{c+ab}{a+b}\)Cho a,b,c>0 và a+b+c=1 Tìm min
Ta có \(a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(a+c\right)\)
\(b+ac=\left(b+a\right)\left(b+c\right)\)
\(c+ab=\left(a+b\right)\left(c+b\right)\)
Đặt \(a+b=x;b+c=y;a+c=z\)=> \(x+y+z=2\)
Khi đó \(P=\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\)
Áp dụng BĐT cosi \(\frac{xy}{z}+\frac{yz}{x}\ge2y\); \(\frac{yz}{x}+\frac{xz}{y}\ge2z\);\(\frac{xy}{z}+\frac{xz}{y}\ge2z\)
Cộng 3 BĐT trên
=> \(P\ge x+y+z=2\)
Vậy MinP=2 khi a=b=c=1/3
Cho a,b,c >0 và a+b+c=1
Tìm Min \(P=\frac{1}{16a}+\frac{1}{9b}+\frac{1}{c}\)
Cho a,b,c>0 thỏa mãn a+b+c=1
Tìm Min: A=\(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}+\frac{1}{9abc}\)
\(A=\text{∑}_{cyc}\frac{a}{a^2+1}+\frac{1}{9abc}=\text{∑}_{cyc}\frac{1}{a+\frac{1}{a}}+\frac{1}{9abc}\)
\(\ge\frac{9}{\text{∑}_{cyc}\left(a+\frac{1}{a}\right)}+\frac{1}{9abc}=P\)
Ta có \(P=\frac{9}{\frac{1}{a+b+c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+\frac{1}{9abc}\)(Vì a + b + c = 1)
\(\ge\frac{9}{\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{9}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+\frac{1}{9abc}\)
\(=\frac{81}{10}.\frac{abc}{ab+bc+ca}+\frac{1}{9abc}\)
\(\Rightarrow P\ge2\sqrt{\frac{3}{ab+bc+ca}}-\frac{21}{10}\ge2\sqrt{\frac{3}{\frac{\left(a+b+c\right)^2}{3}}}-\frac{21}{10}=\frac{39}{10}\)
\(\Rightarrow A\ge P\ge\frac{39}{10}\)
Dấu "=" khi và chỉ khi a = b = c = \(\frac{1}{3}\)
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:
\(A\ge\frac{9}{7}+3=\frac{30}{7}\)
cho a,b,c>0: \(\frac{1}{a+2}+\frac{3}{b+4}=< \frac{c+1}{c+3}\) tìm min Q=(a+1)(b+1)(c+1)
Ta có:
\(\frac{1}{a+2}+\frac{3}{b+4}\le1-\frac{2}{c+3}\)
\(\Rightarrow1-\frac{1}{a+2}\ge\frac{3}{b+4}+\frac{2}{c+3}\ge2\sqrt{\frac{6}{\left(b+4\right)\left(c+3\right)}}\)
\(\Leftrightarrow\frac{a+1}{a+2}\ge2\sqrt{\frac{6}{\left(b+4\right)\left(c+3\right)}}\left(1\right)\)
Tương tự : \(1-\frac{3}{b+4}\ge\frac{1}{a+2}+\frac{2}{c+3}\ge2\sqrt{\frac{2}{\left(a+2\right)\left(c+3\right)}}\Leftrightarrow\frac{b+1}{b+4}\ge2\sqrt{\frac{2}{\left(a+2\right)\left(c+3\right)}}\left(2\right)\)
và \(\frac{c+1}{c+3}\ge2\sqrt{\frac{3}{\left(a+2\right)\left(b+4\right)}}\left(3\right)\)
Từ 1,2,3 ta có:
\(\frac{a+1}{a+2}.\frac{b+1}{b+4}.\frac{c+1}{c+3}\ge\frac{48}{\left(a+2\right)\left(b+4\right)\left(c+3\right)}\Leftrightarrow Q\ge48\)
Vậy Min Q =48 khi a=1,b=5,c=3
a) Cho a,b>0, a+b=<1.Tìm Min của A = \(^{\left(a+\frac{1}{a}\right)^2}\)+ \(^{\left(b+\frac{1}{b}\right)^2}\)
b) Cho a,b,c >0, a+b+c =<1,5. Tìm Min của B= \(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\)
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
Camon bạn!!! Nhưng bạn đọc sai đề r !! ^.^
Cho \(a,b,c>0;a+b+c\le1\). tìm min của \(S=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)