Cho các số dương a,b,c thoả mãn a>b. CMR\(\sqrt{a+c}\)-\(\sqrt{a}\) <\(\sqrt{b+c}\)-\(\sqrt{b}\)
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\(\sqrt{a+c}-\sqrt{a}< \sqrt{b+c}-\sqrt{b}\)
\(\Leftrightarrow\sqrt{a+c}+\sqrt{b}< \sqrt{b+c}+\sqrt{a}\)
\(\Leftrightarrow\left(\sqrt{a+c}+\sqrt{b}\right)^2< \left(\sqrt{b+c}+\sqrt{a}\right)^2\)
\(\Leftrightarrow a+b+c+2\sqrt{ab+bc}< a+b+c+2\sqrt{ab+ac}\)
\(\Leftrightarrow2\sqrt{ab+bc}< 2\sqrt{ab+ac}\Leftrightarrow\sqrt{ab+bc}< \sqrt{ab+ac}\)(đúng vs a>b) .Vậy bđt cần cm đúng
Ta có : \(b=\dfrac{c+a}{2}\Rightarrow2b=c+a\Rightarrow a-b=b-c\)
Dó đó : \(P=\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}-\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}-\sqrt{c}\right)}\right]\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{a-b}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{b-c}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\) Vì \(\left(a-b=b-c\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}+\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\dfrac{\sqrt{a}-\sqrt{c}}{b-c}\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\dfrac{a-c}{a-b}=\dfrac{a-c}{a-\dfrac{a+c}{2}}=\dfrac{a-c}{\dfrac{2a-a-c}{2}}=\dfrac{a-c}{\dfrac{a-c}{2}}=2\)
Lời giải:
Áp dụng BĐT AM-GM:
$\text{VT}=\sqrt{ab+c(a+b+c)}+\sqrt{bc+a(a+b+c)}+\sqrt{ca+b(a+b+c)}$
$=\sqrt{(c+a)(c+b)}+\sqrt{(a+b)(a+c)}+\sqrt{(b+a)(b+c)}$
$\leq \frac{c+a+c+b}{2}+\frac{a+b+a+c}{2}+\frac{b+a+b+c}{2}$
$=2(a+b+c)=2$
Ta có đpcm.
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
\(a-1=-b-c< 0\)trái ĐKXĐ, đề sai , phải đổi 1-a, 1-b, 1-c
- Theo BĐT Cauchy ta có:
\(\sqrt{a.1}\le\dfrac{a+1}{2}\)
\(\sqrt{b.1}\le\dfrac{b+1}{2}\)
\(\sqrt{c.1}\le\dfrac{c+1}{2}\)
\(\sqrt{ab}\le\dfrac{a+b}{2}\)
\(\sqrt{bc}\le\dfrac{b+c}{2}\)
\(\sqrt{ca}\le\dfrac{c+a}{2}\)
\(\Rightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\dfrac{3\left(a+b+c\right)+3}{2}=\dfrac{3.3+3}{2}=6\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Mà ta có: \(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=6\)
\(\Rightarrow a=b=c=1\)
\(M=\dfrac{a^{30}+b^4+c^{1975}}{a^{30}+b^4+c^{2023}}=\dfrac{1^{30}+1^4+1^{1975}}{1^{30}+1^4+1^{2023}}=1\)