Cho a,b,c>0 và a+b+c=1. CMR: \(\frac{a}{a+b^2}+\frac{b}{b+c^2}+\frac{c}{c+a^2}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
1) Cho a,b,c>0 tm a+b+c=3. Cmr \(\frac{1}{2+a^2+b^2}+\frac{1}{2+b^2+c^2}+\frac{1}{2+c^2+a^2}\le\frac{3}{4}\)
2) Cho a,b,c>0 tm a^2+b^2+c^2 bé hơn hoặc bằng abc. Cmr \(\frac{a}{a^2+bc}+\frac{b}{b^2+ca}+\frac{c}{c^2+ab}\le\frac{1}{2}\)
3) Cho a,b,c>0 tm a+b+c<=3. Cmr \(\frac{ab}{\sqrt{3+c}}+\frac{bc}{\sqrt{3+a}}+\frac{ca}{\sqrt{3+b}}\le\frac{3}{2}\)
4) Cho a,b,c>0 tm a+b+c=2. Cmr \(\frac{a}{\sqrt{4a+3bc}}+\frac{b}{\sqrt{4b+3ca}}+\frac{c}{\sqrt{4c+3ab}}\le1\)
5) Cho a,b,c>0. Cmr \(\sqrt{\frac{a^3}{5a^2+\left(b+c\right)^2}}+\sqrt{\frac{b^3}{5b^2+\left(c+a\right)^2}}+\sqrt{\frac{c^3}{5c^2+\left(a+b\right)^2}}\le\sqrt{\frac{a+b+c}{3}}\)
6) Cho a,b,c>0. Cmr \(\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}+\frac{b^2}{\left(2b+a\right)\left(2b+c\right)}+\frac{c^2}{\left(2c+a\right)\left(2c+b\right)}\le\frac{1}{3}\)
Giúp mình với nhé các bạn
cho a,b,c>0 thỏa mãn a+b+c=1
Cmr: \(\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Cho a,b,c >0 CMR:
\(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\le\frac{1}{4}\left(\frac{4}{a}+\frac{5}{b}+\frac{3}{c}\right)\)
Xét hiệu \(VP-VT=\frac{1}{4}\left(\frac{4}{a}+\frac{5}{b}+\frac{3}{c}\right)-\left(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\right)\)
\(=\frac{3a^3b^2+5a^3c^2+3a^2b^3-9a^2b^2c-7a^2bc^2+5a^2c^3+3ab^3c-8ab^2c^2-3abc^3+4b^3c^2+4b^2c^3}{4abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Dễ thấy: \(a;b;c>0\) nên cần chứng minh
\(3a^3b^2+5a^3c^2+3a^2b^3-9a^2b^2c-7a^2bc^2+5a^2c^3+3ab^3c-8ab^2c^2-3abc^3+4b^3c^2+4b^2c^3\ge0\)
\(\Leftrightarrow\frac{1}{2}\left(8a^3+5a^2b+3a^2c-4ab^2-4ac^2-b^3+3b^2c+5bc^2+c^3\right)\left(b-c\right)^2+\frac{1}{2}\left(3a^2c-2a^3-5a^2b+4ab^2+4ac^2+7b^3+3b^2c-5bc^2-c^3\right)\left(c-a\right)^2+\frac{1}{2}\left(2a^3+5a^2b-3a^2c+4ab^2+4ac^2+b^3-3b^2c+5bc^2+9c^3\right)\left(a-b\right)^2\ge0\)
Minh dung phuong phap bieu doi tuong duong thanh tong cac binh phuong do ban nhung cac nay khong hay cho lam.
a) cho x,y dương. CMR: \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)
b) cho a+b+c=1 CMR: \(\frac{a}{a+b^2}+\frac{b}{b+c^2}+\frac{c}{c+a^2}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
a/ \(\Leftrightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2+y^2-2xy\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Vậy BĐT đã cho đúng
b/ \(\frac{a}{a+b^2}=\frac{a}{a\left(a+b+c\right)+b^2}=\frac{a}{a^2+b^2+a\left(b+c\right)}\le\frac{a}{2ab+a\left(b+c\right)}=\frac{1}{b+b+b+c}\)
\(\Rightarrow\frac{a}{a+b^2}=\frac{1}{b+b+b+c}\le\frac{1}{16}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{16}\left(\frac{3}{b}+\frac{1}{c}\right)\)
Tương tự: \(\frac{b}{b+c^2}\le\frac{1}{16}\left(\frac{3}{c}+\frac{1}{a}\right)\) ; \(\frac{c}{c+a^2}\le\frac{1}{16}\left(\frac{3}{a}+\frac{1}{c}\right)\)
Cộng vế với vế:
\(VT\le\frac{1}{16}\left(\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
Cho a,b,c>0 CMR:\(\frac{a}{3a^2+2b^2+c^2}+\frac{b}{3b^2+2c^2+a^2}+\frac{c}{3c^2+2a^2+b^2}\le\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
cho a,b,c >0
CMR:\(\frac{a}{3a^2+2b^2+c^2}+\frac{b}{3b^2+2c^2+a^2}+\frac{c}{3c^2+2a^2+b^2}\le\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
cho a,b,c>0,abc=1
cmr \(\frac{1}{\left(a+1\right)^2+b^2+1}+\frac{1}{\left(b+1\right)^2+c^2+1}+\frac{1}{\left(c+1\right)^2+a^2+1}\le\frac{1}{2}\)
Đầu tiên ta có:
\(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\)
\(=\frac{1}{ab+a+1}+\frac{1}{\frac{1}{a}+b+1}+\frac{1}{\frac{1}{b}+\frac{1}{ab}+1}\)
\(=\frac{1}{ab+a+1}+\frac{a}{1+ab+a}+\frac{ab}{a+1+ab}=1\)
Quay lại bài toán ta có:
\(\frac{1}{\left(a+1\right)^2+b^2+1}=\frac{1}{a^2+b^2+2a+2}\le\frac{1}{2\left(ab+a+1\right)}\)
Tương tự ta có:
\(\hept{\begin{cases}\frac{1}{\left(b+1\right)^2+c^2+1}\le\frac{1}{2\left(bc+b+1\right)}\\\frac{1}{\left(c+1\right)^2+a^2+1}\le\frac{1}{2\left(ca+c+1\right)}\end{cases}}\)
Từ đó suy ra
\(\frac{1}{\left(a+1\right)^2+b^2+1}+\frac{1}{\left(b+1\right)^2+c^2+1}+\frac{1}{\left(c+1\right)^2+a^2+1}\)
\(\le\frac{1}{2}.\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)=\frac{1}{2}\)
Câu hỏi của Nguyễn Trọng Kiên - Toán lớp 9 - Học toán với OnlineMath
cho a,b,c>0
CMR:
\(\left(a+b+\frac{1}{2}\right)^2+\left(b+c+\frac{1}{2}\right)^2+\left(c+a+\frac{1}{2}\right)^2\ge4\left(\frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{b}+\frac{1}{c}}+\frac{1}{\frac{1}{c}+\frac{1}{a}}\right)\)
Áp dụng BĐT Cauchy ta có : \(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}}\Rightarrow\frac{1}{\frac{1}{a}+\frac{1}{b}}\le\frac{\sqrt{ab}}{2}\)
Thiết lập tương tự và thu lại ta có :
\(\Rightarrow VP\le4\left(\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2}\right)=2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(1\right)\)
Áp dụng BĐT Cauchy ta có : \(a+b\ge2\sqrt{ab}\)
\(\Rightarrow\left(a+b+\frac{1}{2}\right)^2\ge\left(2\sqrt{ab}+\frac{1}{2}\right)^2\ge2.2\sqrt{ab}.\frac{1}{2}=2\sqrt{ab}\)
Thiết lập tương tự và thu lại ta có ;
\(\Rightarrow VT\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(2\right)\)
Từ (1) và (2) suy ra
\(VT\ge VP\)
\(\Rightarrowđpcm\)
Chúc bạn học tốt !!!
cho a,b,c>0 . CMR :
\(\left(a+b+\frac{1}{2}\right)^2+\left(b+c+\frac{1}{2}\right)^2+\left(c+a+\frac{1}{2}\right)^2\ge4\left(\frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{b}+\frac{1}{c}}+\frac{1}{\frac{1}{c}+\frac{1}{a}}\right)\)
Áp dụng bđt Cauchy ta có : \(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}}\Rightarrow\frac{1}{\frac{1}{a}+\frac{1}{b}}\le\frac{\sqrt{ab}}{2}\)
Thiết lập tương tự và thu lại ta có :
\(\Rightarrow VP\le4\left(\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2}\right)=2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(1\right)\)
Áp dụng bđt Cauchy ta cso :
\(a+b\ge2\sqrt{ab}\)
\(\Rightarrow\left(a+b+\frac{1}{2}\right)^2\ge\left(2\sqrt{ab}+\frac{1}{2}\right)^2\ge2.2\sqrt{ab}.\frac{1}{2}=2\sqrt{ab}\)
Thiết lập tương tự và thu lại ta có :
\(\Rightarrow VT\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(2\right)\)
Từ (1) và (2)
\(VT\ge VP\)
\(\Rightarrowđpcm\)
Chúc bạn học tốt !!!