lú rùi vậy cũng sai :(
\(BDT\Leftrightarrow\sqrt{\dfrac{c}{b}.\dfrac{a-c}{a}}+\sqrt{\dfrac{c}{a}.\dfrac{b-c}{b}}\le1\)
Áp dụng BĐT AM-GM ta có:
\(VT\le\dfrac{\dfrac{c}{b}+\dfrac{a-c}{a}}{2}+\dfrac{\dfrac{c}{a}+\dfrac{b-c}{b}}{2}=1\)
Cho a,b,c > 0 thỏa abc=1.Chứng minh :
\(P=\dfrac{1}{\sqrt{a\left(1+b\right)}}+\dfrac{1}{\sqrt{b\left(1+c\right)}}+\dfrac{1}{\sqrt{c\left(1+a\right)}}>2\)
cho a , b , c > 0 thỏa mãn \(a+b+c+\sqrt{abc}=4\)
Tính giá trị : \(p=\sqrt{a\left(4-b\right)\left(4-c\right)+b\left(4-c\right)\left(4-a\right)-c\left(4-a\right)\left(4-b\right)}-\sqrt{abc}\)
CM bất đẳng thức sau:
a, Cho a>c , b>c , c>0
CM: \(\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{ab}\)
b, CM
\(\dfrac{2005}{\sqrt{2006}}+\dfrac{2006}{\sqrt{2005}}>\sqrt{2005}+\sqrt{2006}\)
help me!!
cho a,b,c >0 ; \(a+b+c=a^2+b^2+c^2=2\)
Tinnsh \(A=a\sqrt{\dfrac{\left(1+b^2\right)\left(1+c^2\right)}{1+a^2}}+b\sqrt{\dfrac{\left(1+a^2\right)\left(1+c^2\right)}{1+b^2}}+c\sqrt{\dfrac{\left(1+a^2\right)\left(1+b^2\right)}{1+c^2}}\)
cho a, b, c \(\ge\) 0, a+b+c=3. tìm Max
K=\(\sqrt{12a+\left(b-c\right)^2}+\sqrt{12b+\left(c-a\right)^2}+\sqrt{12c+\left(a-b\right)^2}\)
Với a,b,c,d thuộc Q thỏa mãn a+b+c+d=0. CMR x=\(\sqrt{\left(ab-cd\right)\left(bc-ad\right)\left(ca-bd\right)}\in Q\)
Cho \(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\) . Chứng minh \(\dfrac{\sqrt{a}}{1+a}+\dfrac{\sqrt{b}}{1+b}+\dfrac{\sqrt{c}}{1+c}=\dfrac{2}{\sqrt{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
1/ cho a,b,c thỏa \(ab+bc+ca\ge11\)
c/m \(\sqrt[3]{a^2+3}+\dfrac{7}{5\sqrt[3]{14}}\sqrt[3]{b^2+3}+\dfrac{\sqrt[3]{9}}{5}\sqrt[3]{c^2+3}\ge\dfrac{23}{5\sqrt[3]{2}}\)
2)cho a,b,c dương thỏa a+b+c=3
c/m \(\left(a^3+b^3+c^3\right)\left(a^2-b^2\right)\left(b^2-c^2\right)\left(c^2-a^2\right)\le\dfrac{729\sqrt{3}}{8}\)
p/s: cách của mik đa phần dùng cô-si (I need another way!!)
ba số dương a,b,c thỏa mãn \(b\ne c,\sqrt{a}+\sqrt{b}\ne\sqrt{c}\) và\(a+b=\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)^2\).chứng minh đẳng thức
\(\dfrac{a+\left(\sqrt{a}-\sqrt{c}\right)^2}{b+\left(\sqrt{b}-\sqrt{c}\right)^2}=\dfrac{\sqrt{a}-\sqrt{c}}{\sqrt{b}-\sqrt{c}}\)
