Cho a,b,c>0 và \(ab+bc+ca\ge\frac{4}{3}\).chứng minh
\(\sqrt{a^2+\frac{1}{\left(b+1\right)^2}}+\sqrt{b^2+\frac{1}{\left(c+1\right)^2}}+\sqrt{c^2+\frac{1}{\left(a+1\right)^2}}\ge\frac{\sqrt{181}}{5}\)
Cho: \(a>0,\) \(b>0\)
Chứng minh rằng: \(\sqrt{a}\left(\sqrt{\frac{a}{b}}-1\right)\ge\sqrt{b}\left(1-\sqrt{\frac{b}{a}}\right)\)
Cho a,b,c > 0 và ab + bc + ca \(\ge\frac{4}{3}\)
Chứng minh :
\(\sqrt{a^2+\frac{1}{\left(b+1\right)^2}}+\sqrt{b^2+\frac{1}{\left(c+1\right)^2}}+\sqrt{c^2+\frac{1}{\left(a+1\right)^2}}\ge\frac{\sqrt{181}}{5}\)
Bài 1: Cho a,b>0. Chứng minh \(\sqrt[3]{\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)}< \sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}\)
Bài 2: Cho a,b>0. Chứng minh \(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\ge\frac{2\sqrt{2}}{\sqrt{a+b}}\)
Bài 3: Cho a,b,c>0. Chứng minh \(\frac{5b^3-a^3}{ab+3b^2}+\frac{5c^3-b^3}{bc+3c^2}+\frac{5a^3-c^3}{ca+3a^2}\le a+b+c\)
Với mọi a, b>0 chứng minh \(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{1}{2}\)
1.Chứng minh \(\sqrt{x^2+xy+y^2}+\sqrt{x^2+xz+z^2}\ge\sqrt{y^2+yz+z^2}\)
2. Cho a,b,c>0. Chứng minh \(\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)\left(\frac{1}{\sqrt[3]{a}}+\frac{1}{\sqrt[3]{b}}+\frac{1}{\sqrt[3]{c}}\right)-\frac{a+b+c}{\sqrt[3]{abc}}\le6\)
3. Cho a,b>0 , n là số nguyên dương. Chứng minh \(\frac{1}{\sqrt[n]{a}}+\frac{1}{\sqrt[n]{b}}\ge2\sqrt[n]{\frac{2}{a+b}}\)
4. Cho a,b,c >0. Chứng minh \(\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ba}\le\frac{a+b+c}{2abc}\)
chứng minh câu đẳng thức
1)\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
2)\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
3)\(\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{2b}{a-b}=1\)(a lớn hơn bằng 0,b lớn hơn bằng 0)
4)\(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\)(a lớn hơn bằng 0,a khác 1)
help me:<<<
Chứng minh rằng \(\frac{1}{2\sqrt[3]{abc}}+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\forall a,b,c>0\)
các bạn làm được ý nào thì làm ý đó nha
1. Cho a,b,c là độ dài 3 cạnh tam giác. Chứng minh:
a) \(\frac{1}{\left(a+b-c\right)^2}+\frac{1}{\left(a-b+c\right)^2}+\frac{1}{\left(b+c-a\right)^2}\ge\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
b) \(\frac{1}{\left(a+b-c\right)^3}+\frac{1}{\left(a-b+c\right)^3}+\frac{1}{\left(b+c-a\right)^3}\ge\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\)
c) \(\frac{1}{\left(a+b-c\right)^{200}}+\frac{1}{\left(a-b+c\right)^{200}}+\frac{1}{\left(b+c-a\right)^{200}}\ge\frac{1}{a^{200}}+\frac{1}{b^{200}}+\frac{1}{c^{200}}\)
d) \(\frac{1}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\sqrt{abc\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)}\)
e) \(a+b+c< \sqrt{a\left(b+c\right)}+\sqrt{b\left(a+c\right)}+\sqrt{c\left(a+b\right)}\)
f) \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}< \sqrt{6}\)
g) \(\sqrt{-a+b+c}+\sqrt{a-b+c}+\sqrt{a+b-c}\le\sqrt{3\left(a+b+c\right)}\)