Lời giải:
Do $a>b>0$ nên $2ab-b^2> 2b^2-b^2=b^2$
$\Rightarrow \sqrt{a^2-b^2}+\sqrt{2ab-b^2}> \sqrt{a^2-b^2}+\sqrt{b^2}=\sqrt{a^2-b^2}+b$
Mà:
$\sqrt{a^2-b^2}+b=\sqrt{(\sqrt{a^2-b^2}+b)^2}$
$=\sqrt{a^2+2b\sqrt{a^2-b^2}}>\sqrt{a^2}=a$
Do đó: $\sqrt{a^2-b^2}+\sqrt{2ab-b^2}>a$ (đpcm)
Cho các số dương a, b, c thỏa mãn: a+b+c=3 và \(M=\sqrt{a^2+2ab+2b^2}+\sqrt{b^2+2bc+2c^2}+\sqrt{c^2+2ca+2a^2}\). CMR: \(M\ge3\sqrt{5}\)
Cho a, b>0. Chứng minh rằng:
a) \(\dfrac{3a^2+2ab+3b^2}{a+b}\ge2\sqrt{2\left(a^2+b^2\right)}\)
b) \(\dfrac{2ab}{a+b}+\sqrt{\dfrac{a^2+b^2}{2}}\ge\sqrt{ab}+\dfrac{a+b}{2}\)
c) \(\dfrac{1}{\left(1+a\right)^2}+\dfrac{1}{\left(1+b\right)^2}\ge\dfrac{1}{1+ab}\)
(4)Bài 1:Với \(\forall\) a>b>0. CMR: a+ \(\frac{1}{b\left(a-b\right)}\ge3\)
(7) Bài 2: Cho a,b,c \(\ne\) 0 .CMR: \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)
(8) Bài 3: Cho a,b,c>0 thõa mãn abc=1
CMR: \(\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
Cho a, b, c>0. CMR: \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Cho a, b, c>0 và a+b+c\(\ge3\)
Cmr:
\(\dfrac{a^2}{a+\sqrt{bc}}+\dfrac{b^2}{b+\sqrt{ac}}+\dfrac{c^2}{c+\sqrt{ab}}\ge\dfrac{3}{2}\)
Cho các số dương a, b, c thỏa mãn: a+b+c=3 và \(M=\sqrt{a^2+2ab+2b^2}+\sqrt{b^2+2bc+2c^2}+\sqrt{c^2+2ca+2a^2}\)
Cho a,b,c > 0 có a+b+c \(\le3\)
CMR : \(\dfrac{a}{\sqrt{2a^2+b^2}+\sqrt{3}}+\dfrac{b}{\sqrt{2b^2+c^2}\sqrt{3}}+\dfrac{c}{\sqrt{2c^2+a^2}+\sqrt{3}}\le\dfrac{\sqrt{3}}{2}\)
Cho a,b,c > 0. CMR:
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\)
Cho a,b,c>0 thỏa a+b+c=4
CMR: \(\sqrt[4]{a^3}+\sqrt[4]{b^3}+\sqrt[4]{c^3}>2\sqrt{2}\)
