Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{a+\sqrt{\left(\sqrt{ac}+\sqrt{ab}\right)^2}}=\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{c}+\sqrt{b}}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\le\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1=VP\)
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 )
^_^
a, cho a,b,c là các số hữu tỉ khác nhau.CMR:\(\sqrt{\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{\left(b-c\right)^2}+\dfrac{1}{\left(a-c\right)^2}}\)
là 1 số hữu tỉ
câu1 : a) A= \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}\)
b) \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
Câu 2 :
a) A= \(\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right).\left(\sqrt{10}-\sqrt{2}\right)\)
b) B= \(\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right).\left(1-\dfrac{2}{a+1}\right)^2\)
\(CHO\:A\:,b,c,\:x,y,z,>0\:VA\dfrac{A}{X}=\dfrac{B}{Y}=\dfrac{C}{Z}\:CM:\:\sqrt{AX}+\sqrt{BY}+\sqrt{CZ\:}=\left(\sqrt{A+b+c\:}\right)\:\left(\sqrt{X+y+z}\right)\)
a, Giải phương trình: 2\(\left(x-\sqrt{2x^2+5x-3}\right)=1+x\left(\sqrt{2x-1}-2\sqrt{x+3}\right)\)
b, Cho ba số thực dương a,b,c thỏa mãn a,b,c=1
Chứng minh rằng:\(\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2}\)
Cho a;b;c >0 thỏa mãn \(a+b+c=\dfrac{1}{abc}\)
Cmr: \(\sqrt{\dfrac{\left(1+b^2c^2\right)\left(1+a^2c^2\right)}{c^2+a^2b^2c^2}}=a+b\)
Giúp em với ạ. Em cảm ơn các anh/chị ạ.
cho các số dương a,b,c thỏa mãn abc=1. chứng minh rằng
\(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c^3}{\left(1+a\right)\left(1+c\right)}\ge\dfrac{3}{4}\)
a,b,c >0 thỏa mãn b≠c ,\(\sqrt{a}\)+\(\sqrt{b}\)≠\(\sqrt{c}\) và a+b=\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2.cmr\)
\(\dfrac{a+(\sqrt{a}-\sqrt{c})^2}{b+\left(\sqrt{b}-\sqrt{c}\right)}\)=\(\dfrac{\sqrt{a}-\sqrt{c}}{\sqrt{b}-\sqrt{c}}\)
cho \(\sqrt{a}+\sqrt{\sqrt{b}+}\sqrt{c}=\sqrt{3}va\sqrt{\left(a+2b\right)\left(a+2c\right)}+\sqrt{\left(b+2a\right)\left(b+2c\right)}+\sqrt{\left(c+2a\right)\left(c+2b\right)}=3\)
tính M=\(\left(2\sqrt{a}+3\sqrt{b}-4\sqrt{c}\right)^2\)
