Sử dụng AM-GM, ta có
\(P=\sum\sqrt{\dfrac{ab}{ab+c}}=\sum\sqrt{\dfrac{ab}{ab+c\left(a+b+c\right)}}=\sum\sqrt{\dfrac{ab}{\left(c+b\right)\left(c+a\right)}}\le\dfrac{1}{2}\sum\dfrac{a}{c+b}+\dfrac{b}{c+a}=\dfrac{3}{2}\)
từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)
ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)
=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)
\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )
^_^
Cho a,b,c dương thỏa mãn \(ab+bc+ca=1\). Tìm GTLN của
\(P=\dfrac{2a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}\)
Cho a,b,c > 0 và a+b +c =1
Tìm Max A = \(\dfrac{ab}{\sqrt{c+ab}}+\dfrac{bc}{\sqrt{a+bc}}+\dfrac{ca}{\sqrt{b+ca}}\)
cho a,b,c>0 và \(a^2+b^2+c^2=1\)
chứng minh rằng \(A=\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}+\sqrt{\dfrac{bc-2a^2}{1+bc-a^2}}+\sqrt{\dfrac{ca+2b^2}{1+ca-b^2}}\ge2+ab+bc+ca\)
a) Cho a , b > 0 CMR : 3(b2+2a2) ≥ (b+2a)2
b) Cho a,b,c > 0 thõa mãn ab+bc+ca = abc
CMR : \(\dfrac{\sqrt{b^2+2a^2}}{ab}+\dfrac{\sqrt{c^2+2b^2}}{bc}+\dfrac{\sqrt{a^2+2c^2}}{ca}\ge\sqrt{3}\)
Rut gon phuong trinh \(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a^6}+\sqrt{b^6}}{a-b}\)
A \(\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
B\(\dfrac{\sqrt{ab}-2b}{\sqrt{a}-\sqrt{b}}\)
C \(\dfrac{2b}{\sqrt{a}-\sqrt{b}}\)
D\(\dfrac{\sqrt{ab}-2a}{\sqrt{a}-\sqrt{b}}\)
Giup minh chon dap an dung
cho a ,b,c duong . CMR :
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}\) >2
Cho a, b, c dương và a + b + c = 2. Tìm GTLN của \(\sqrt{2a+bc}+\sqrt{2c+ab}+\sqrt{2b+ac}\)
cho 3 số thực a,b,c >0 thỏa mãn \(a^2+b^2+c^2=3\) ,chứng minh:
\(\dfrac{1}{4-\sqrt{ab}}+\dfrac{1}{4-\sqrt{bc}}+\dfrac{1}{4-\sqrt{ca}}\le1\)
dấu đẳng thức xảy ra khi nào ?
