\(choa,b,c>0\) và \(ab+bc+ca=3.cmr:\sqrt{a^6+b^6+1}+\sqrt{b^6+c^6+a}+\sqrt{c^6+a^6+a}\ge3\sqrt{3}\)\(\)
Với a,b,c > 0 thỏa mãn ab + bc + ca = 3 . CMR:
\(\sqrt{a^6+b^6+1}+\sqrt{b^6+c^6+1}+\sqrt{c^6+a^6+1}\ge3\sqrt{3}\)
Chậc -.- ai ngờ bài này lại dễ vậy .... Cứ chứng minh đủ kiểu hóa ra dùng Cô-si là xong .... nghĩ xa quá XD
Áp dụng bđt Cô-si cho 3 số dương ta được
\(\sqrt{a^6+b^6+1}\ge\sqrt{3\sqrt[3]{a^6.b^6.1}}=ab\sqrt{3}\)
C/m tương tự \(\sqrt{b^6+c^6+1}\ge bc\sqrt{3}\)
\(\sqrt{c^6+a^6+1}\ge ac\sqrt{3}\)
Cộng 3 bđt trên lại ta được
\(VT\ge\left(ab+bc+ca\right)\sqrt{3}=3\sqrt{3}\)
Dấu "=" xảy ra <=> a = b= c = 1
Vậy ..........
Cho a, b, c thỏa mãn: \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7;a+b+c=23;\sqrt{abc}=3\). Tính giá trị của biểu thức: \(H=\dfrac{1}{\sqrt{ab}+\sqrt{c}-6}+\dfrac{1}{\sqrt{bc}+\sqrt{a}-6}+\dfrac{1}{\sqrt{ca}+\sqrt{b}-6}\)
Cho các số thực dương a,b,c thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\). CMR:
\(\frac{a+b}{\sqrt{ab+c}}+\frac{b+c}{\sqrt{bc+a}}+\frac{c+a}{\sqrt{ca+b}}\ge3\sqrt[6]{abc}\)
Giải:
\(GT\Leftrightarrow ab+bc+ca\ge abc\)
\(\Rightarrow ab\le\frac{ab+bc+ca}{c}\)
\(\Rightarrow\frac{a+b}{\sqrt{ab+c}}\ge\frac{a+b}{\sqrt{\frac{ab+bc+ca}{c}+c}}=\frac{\left(a+b\right)\sqrt{c}}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
Tương tự rồi cộng lại: \(VT\ge\frac{\left(a+b\right)\sqrt{c}}{\sqrt{\left(c+a\right)\left(c+b\right)}}+\frac{\left(b+c\right)\sqrt{a}}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{\left(c+a\right)\sqrt{c}}{\sqrt{\left(b+a\right)\left(b+c\right)}}\)\(\ge3\sqrt[3]{\sqrt{abc}}=3\sqrt[6]{abc}\)
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Cho a,b,c >0 thỏa mãn : \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=6\) 6
cmr \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}>=3\)
Mọi người ơi chỉ =6 thôi nha k phải 66 đâu
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{b+c+c+a+a+b}=\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\)
\(\ge\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\frac{6}{2}=3\)(BĐT \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
Dấu "=" xảy ra khi \(a=b=c=2\)
Cho 3 số dương \(ab\sqrt{ab}+bc\sqrt{bc}+ca\sqrt{ca}=1\)Tìm min P=\(\frac{a^6}{a^3+b^3}+\frac{b^6}{b^3+c^3}+\frac{c^6}{c^3+a^3}\)
Chờ a,b,c>0 và a+b+c=6
CMR \(\frac{ab}{\sqrt{c^2+3}}+\frac{bc}{\sqrt{a^2+3}}+\frac{ca}{\sqrt{b^2+3}}\le\frac{3}{2}\)
bn xem lại cái đề nhé, với a = b = c = 2 thì ko đúng đâu
Cho a,b,c >0 tm abc=1, C/m
\(\dfrac{1}{\sqrt{a^5+b^2+ab+6}}+\dfrac{1}{\sqrt{b^5+c^2+bc+6}}+\dfrac{1}{\sqrt{c^5+a^2+ca+6}}\le1\)
\(a^5+b^2+ab+6\ge3a^2b+6\)
\(\Rightarrow P\le\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{\sqrt{a^2b+2}}+\dfrac{1}{\sqrt{b^2c+2}}+\dfrac{1}{\sqrt{c^2a+2}}\right)\le\sqrt{\dfrac{1}{a^2b+2}+\dfrac{1}{b^2c+2}+\dfrac{1}{c^2a+2}}=\sqrt{Q}\)
\(Q=\dfrac{c}{a+2c}+\dfrac{a}{b+2a}+\dfrac{b}{c+2b}=\dfrac{1}{2}\left(1-\dfrac{a}{a+2c}+1-\dfrac{b}{b+2a}+1-\dfrac{c}{c+2b}\right)\)
\(Q=\dfrac{3}{2}-\dfrac{1}{2}\left(\dfrac{a^2}{a^2+2ac}+\dfrac{b^2}{b^2+2ab}+\dfrac{c^2}{c^2+2bc}\right)\)
\(Q\le\dfrac{3}{2}-\dfrac{1}{2}\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1\)
\(\Rightarrow P\le\sqrt{1}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
cho a, b, c thỏa mãn \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7;a+b+c=23;\sqrt{abc}=3\)
tính giá trị biểu thức \(A=\frac{1}{\sqrt{ab}+\sqrt{c}-6}+\frac{1}{\sqrt{bc}+\sqrt{a}-6}+\frac{1}{\sqrt{ca}+\sqrt{b}-6}\)
Câu hỏi của hoàng thị huyền trang - Toán lớp 9 - Học toán với OnlineMath
Em tham khảo nhé!
cho a, b, c >0 tm \(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=6\)
CMR \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{a+c}\ge3\)
Cái này không khó :v
Áp dụng BĐT Cauchy-Schwarz dạng Engel, ta có:
\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{a+c}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Face khác ;v, theo AM-GM, ta có
\(\dfrac{a+b+c}{2}\ge\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\dfrac{6}{2}=3\)
Vậy ta có đpcm. Đẳng thức xảy ra khi a=b=c=2