Cho a,b,c la cac so duong a+b+c=3
Chung minh:\(a^5+b^5+c^5+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge6\)
cho cac so thuc duong a b c thoa a^2+b^2+c^2>=3 chung minh
\(\frac{\left(a+1\right)\left(b+2\right)}{\left(b+1\right)\left(b+5\right)}+\frac{\left(b+1\right)\left(c+2\right)}{\left(c+1\right)\left(c+5\right)}+\frac{\left(c+1\right)\left(a+2\right)}{\left(a+1\right)\left(a+5\right)}\ge\frac{3}{2}\)
Ta có đánh giá \(\frac{b+2}{\left(b+1\right)\left(b+5\right)}\ge\frac{3}{4\left(b+2\right)}\)
Thật vậy, BĐT trên tương đương:
\(4\left(b+2\right)^2\ge3\left(b+1\right)\left(b+5\right)\)
\(\Leftrightarrow b^2-2b+1\ge0\Leftrightarrow\left(b-1\right)^2\ge0\) (luôn đúng)
\(\Rightarrow\frac{\left(a+1\right)\left(b+2\right)}{\left(b+1\right)\left(b+5\right)}\ge\frac{3\left(a+1\right)}{4\left(b+2\right)}\)
Tương tự và cộng lại: \(P\ge\frac{3}{4}\left(\frac{a+1}{b+2}+\frac{b+1}{c+2}+\frac{c+1}{a+2}\right)\)
\(P\ge\frac{3}{4}\left(\frac{\left(a+1\right)^2}{ab+2a+b+2}+\frac{\left(b+1\right)^2}{bc+2b+c+2}+\frac{\left(c+1\right)^2}{ca+2c+a+2}\right)\)
\(P\ge\frac{3}{4}.\frac{\left(a+b+c+3\right)^2}{ab+bc+ca+3a+3b+3c+6}\)
\(P\ge\frac{3}{4}.\frac{a^2+b^2+c^2+2ab+2bc+2ca+6a+6b+6c+9}{ab+bc+ca+3a+3b+3c+6}\)
\(P\ge\frac{3}{4}.\frac{2ab+2bc+2ca+6a+6b+6c+12}{ab+bc+ca+3a+3b+3c+6}=\frac{3}{4}.2=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho a,b,c,d la cac so duong sao cho a+b+c = 1 . Chung minh \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}\ge\frac{1}{2}\)
Cho a,b,c la cac so duong sao cho a+b+c = 1
Chung minh \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}>=\frac{1}{2}\)
cmtt \(\frac{b^2}{a+c}+\frac{a+c}{4}\ge b\)
\(\frac{c^2}{a+b}+\frac{a+b}{4}\ge c\)
\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}+\frac{a+b+b+c+c+a}{4}\ge a+b+c\)
\(A+\frac{1}{2}\ge1\)
\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2}{4}}=a\)
cmtt
A+1/2\(\ge1\Rightarrow A\ge\frac{1}{2}\)
A là biểu thức bên trái nha
1. Cho a,b la 2 so duong thoa a+b<=1.chung minh rang \(6b+\frac{1}{3a}+\frac{4}{b}\ge11\).
2. cho a,b,c la cac so nguyen duong sao cho (a-b).(a-c).(b-c)=a+b+c
a. chung minh rang a+b+c chia het cho 2
b. Tim gia tri nho nhat cua M=a+b+c
cho a, b, c la cac so duong thoa man a\(a^2+b^2+c^2=3\) . Chung minh rang : \(\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2-c}>=3\)
???? là sao vừa lớn vừa bằng đó
duyệt đi
cho a, b, c la cac so thuc duong thoa man a + b + c =abc chung minh rang :
\(\frac{1}{a^2\left(1+bc\right)}+\frac{1}{b^2\left(1+ac\right)}+\frac{1}{c^2\left(1+ab\right)}\le\frac{1}{4}\)
\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(VT=\frac{x^2yz}{1+yz}+\frac{xy^2z}{1+zx}+\frac{xyz^2}{1+xy}=\frac{x^2yz}{xy+yz+yz+zx}+\frac{xy^2z}{xy+zx+yz+zx}+\frac{xyz^2}{xy+yz+xy+zx}\)
\(VT\le\frac{1}{4}\left(\frac{x^2yz}{xy+yz}+\frac{x^2yz}{yz+zx}+\frac{xy^2z}{xy+zx}+\frac{xy^2z}{yz+zx}+\frac{xyz^2}{xy+yz}+\frac{xyz^2}{xy+zx}\right)\)
\(VT\le\frac{1}{4}\left(\frac{x^2y}{x+y}+\frac{xy^2}{x+y}+\frac{y^2z}{y+z}+\frac{yz^2}{y+z}+\frac{x^2z}{x+z}+\frac{xz^2}{x+z}\right)\)
\(VT\le\frac{1}{4}\left(xy+yz+zx\right)=\frac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)
cho a,b,c > 0 và a+b+c =3. chứng minh \(a^5+b^5+c^5+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge6\)
áp dụng Cô-si ta có:
\(a^5+\frac{1}{a}+1+1\ge4\sqrt[4]{a^5.\frac{1}{a}.1.1}=4a\)
\(b^5+\frac{1}{b}+1+1\ge4\sqrt[4]{b^5.\frac{1}{b}.1.1}=4b\)
\(c^5+\frac{1}{c}+1+1\ge4\sqrt[4]{c^5.\frac{1}{c}.1.1}=4c\)
\(\Rightarrow a^5+b^5+c^5+1+1+1+1+1+1\ge4a+4b+4c\)
\(\Leftrightarrow a^5+b^5+c^5\ge4\left(a+b+c\right)-6=4.3-6=6\)
Dấu = xảy ra khi a=b=c=1
Vẫn áp dụng cô si nhưng lần này sẽ khác cách của Thành:
Áp dụng BĐT Côsi,ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
Suy ra \(VT\ge a^5+b^5+c^5+3\sqrt[3]{\frac{1}{abc}}\)
Suy ra \(VT+1+1\ge a^5+b^5+c^5+1+1+3\sqrt[3]{\frac{1}{abc}}\) (1)
Áp dụng Côsi,ta có: \(a^5+b^5+c^5+1+1\ge5\sqrt[5]{1a^5b^5c^51}=5abc\)(2)
Từ (1) và (2) suy ra \(VT+1+1\ge5abc+3\sqrt[3]{\frac{1}{abc}}\)
\(VT\ge5abc+3\sqrt[3]{\frac{1}{abc}}-2\).Ta cần chứng minh \(5abc+3\sqrt[3]{\frac{1}{abc}}-2\ge6\Leftrightarrow5abc+3\sqrt[3]{\frac{1}{abc}}\ge8\) (3)
Thật vậy ta có: \(\sqrt[3]{abc}\le\frac{a+b+c}{3}\Rightarrow abc\ge\frac{a+b+c}{3}\).Áp dụng vào,ta có:
\(abc\ge\frac{a+b+c}{3}=1\) (do a + b + c = 3).
Thay vào (3),ta có:\(5abc+3\sqrt[3]{\frac{1}{abc}}\ge8\) suy ra \(5abc+3\sqrt[3]{\frac{1}{abc}}-2\ge6\) suy ra đpcm
Gia su a, b, c la cac so duong, chung minh rang: \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2\)
dùng bđt cauchy chứng minh biểu thức trên >=2 rồi chứng minh dấu = không xảy ra
Cho a,b.c la cac so duong va abc = 1
Chung minh rang \(\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\)