cho a, b, c la cac so thuc duong thoa man a + b + c =abc chung minh rang :
\(\frac{1}{a^2\left(1+bc\right)}+\frac{1}{b^2\left(1+ac\right)}+\frac{1}{c^2\left(1+ab\right)}\le\frac{1}{4}\)
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 la cac so nguyen duong thoa man: abc=1. CMR
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\)
bài này chứng minh bài toán phụ, khá là phức tạp, trình bày ra chắc chết quá
bài này mình thấy tren mạng đăng lên đó, có kết quả nhưng ko copy được
Bài này bạn xem lại trong chtt ấy! Mình giải bài này rồi, giải bằng miệng cho nhanh.
cho 2 so thuc a,b thoa man a>1va b>1 chung minh rang\(\frac{a^3+b^3-\left(a^2+b^2\right)}{\left(a-1\right)\left(b-1\right)}\)
\(A=\frac{a^3+b^3-\left(a^2+b^2\right)}{\left(a-1\right)\left(b-1\right)}=\frac{a^2\left(a-1\right)+b^2\left(b-1\right)}{\left(a-1\right)\left(b-1\right)}=\frac{a^2}{b-1}+\frac{b^2}{a-1}\)
(chơi 3 cách luôn cho máu :3)
Cách 1, Áp dụng Svacxơ đc
\(A=\frac{a^2}{b-1}+\frac{b^2}{a-1}\ge\frac{\left(a+b\right)^2}{a+b-2}=\frac{t^2}{t-2}\left(t=a+b>2\right)\)
Ta luôn có \(\frac{t^2}{t-2}\ge8\left(1\right)\)thật vậy
\(\left(1\right)\Leftrightarrow t^2\ge8t-16\Leftrightarrow t^2-8t+16\ge0\Leftrightarrow\left(t-4\right)^2\ge0\left(True\right)\)
=> Đpcm
Cách 2, \(A=\frac{a^2}{b-1}+\frac{b^2}{a-1}\ge2\sqrt{\frac{a^2.b^2}{\left(b-1\right)\left(a-1\right)}}=2.\frac{a}{\sqrt{a-1}}.\frac{b}{\sqrt{b-1}}\)
Ta đi c/m \(\frac{a}{\sqrt{a-1}}\ge2\left(#\right)\)thật vậy
\(\left(#\right)\Leftrightarrow a\ge2\sqrt{a-1}\Leftrightarrow a^2\ge4a-4\Leftrightarrow a^2-4a+4\ge0\Leftrightarrow\left(a-2\right)^2\ge0\left(true\right)\)
=> (#) đúng
tương tự\(\frac{b}{\sqrt{b-1}}\ge2\)
\(\Rightarrow A\ge2.2.2=8\)(Đpcm)
Cách 3 , \(A=\frac{a^2}{b-1}+\frac{b^2}{a-1}=\frac{\left(a-1+1\right)^2}{b-1}+\frac{\left(b-1+1\right)^2}{a-1}\)
\(=\frac{\left(a-1\right)^2+2\left(a-1\right)+1}{b-1}+\frac{\left(b-1\right)^2+2\left(b-1\right)+1}{a-1}\)
\(=\frac{\left(a-1\right)^2}{b-1}+\frac{2\left(a-1\right)}{b-1}+\frac{1}{b-1}+\frac{\left(b-1\right)^2}{a-1}+\frac{2\left(b-1\right)}{a-1}+\frac{1}{a-1}\)
\(=\left[\frac{\left(a-1\right)^2}{b-1}+\frac{\left(b-1\right)^2}{a-1}\right]+2\left(\frac{a-1}{b-1}+\frac{b-1}{a-1}\right)+\left(\frac{1}{b-1}+\frac{1}{a-1}\right)\)
\(\ge2\sqrt{\frac{\left(a-1\right)^2.\left(b-1\right)^2}{\left(b-1\right)\left(a-1\right)}}+2.2\sqrt{\frac{a-1}{b-1}.\frac{b-1}{a-1}}+\frac{2}{\sqrt{\left(a-1\right)\left(b-1\right)}}\)
\(=2\sqrt{\left(a-1\right)\left(b-1\right)}+\frac{2}{\sqrt{\left(a-1\right)\left(b-1\right)}}+4\)
\(\ge2\sqrt{2\sqrt{\left(a-1\right)\left(b-1\right)}.\frac{2}{\sqrt{\left(a-1\right)\left(b-1\right)}}}+4\)
\(=2.2+4=8\)
Dấu "=" xảy ra tại a = b = 2
cho a , b, c la cac so thuc duong thoa man he thuc a+b+c=6abc
Chung minh rang \(\dfrac{bc}{a^3\left(c+2b\right)}+\dfrac{ac}{b^3\left(a+2c\right)}+\dfrac{ab}{c^3\left(b+2a\right)}\ge2\)
cho 3 so thuc a,b,c thoa man \(\frac{1}{a+2}+\frac{3}{b+4}\le\frac{c+1}{c+3}\)
tim min cua \(q=\left(a+1\right)\left(b+1\right)\left(c+1\right)\)
\(\frac{1}{a+2}+\frac{3}{b+4}+\frac{2}{c+3}\le1\Leftrightarrow x+y+z\le1\)
\(Q=\left(\frac{1}{x}-1\right)\left(\frac{3}{y}-3\right)\left(\frac{2}{z}-2\right)=\frac{6\left(1-x\right)\left(1-y\right)\left(1-z\right)}{xyz}\ge\frac{6\left(y+z\right)\left(x+z\right)\left(x+y\right)}{xyz}\ge6.2.2.2=48\)
Min Q = 48 khi x =y=z = 1/3 => a =1 ; b =5; c =3
cho a, b, c, d la 4 so nguyen duong thoa man: b= \(\frac{a+c}{2}va\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{b}+\frac{1}{d}\right)\)
chung minh: \(\frac{a}{b}=\frac{c}{d}\)
Cho 2 so thuc a, b thoa man dieu kien ab= 1, a+ b\(\ne\)0. Tinh gia tri bieu thuc :
P= \(\frac{1}{\left(a+b\right)^3}\left(\frac{1}{a^3}+\frac{1}{b^3}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{6}{\left(a+b\right)^3}\left(\frac{1}{a}+\frac{1}{b}\right)\)
1.tìm các nghiem nguyen cua phuong trinh: 54x^3+1=y^3
2.cho x+y=1 và xy khac 0.chung mih \(\frac{x}{y^3-1}+\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)
3.cho a,b,c la cac so thuc duong.chung minh :\(\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)^2+\frac{14abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge4\)
Câu 2 thế y = 1 - x rồi quy đồng như bình thường là ra bn nhé
Cho 3 so thuc a,b,c khong am thỏa mãn (a+b)(b+c)(c+a)>0.Chứng minh rằng
\(\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(a+c\right)^2}\ge\)\(\frac{9}{4\left(ab+bc+ac\right)}\)