Cho \(a,b,c>0.\) Chứng minh rằng \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Cho a , b , c > 0 . Chứng minh rằng :
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
Tương tự: \(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c}\) ; \(\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Cộng vế:
\(VT\ge\dfrac{2a+2b+2c}{a+b+c}=2\)
Dấu "=" ko xảy ra nên \(VT>2\)
cho a,b,c >0 chứng minh rằng
\(\sqrt{\dfrac{a+b}{c}}+\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}>=2\left(\sqrt{\dfrac{c}{a+b}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{a}{b+c}}\right)\)
Lời giải:
Đặt \(\left ( \sqrt{\frac{a}{b+c}},\sqrt{\frac{b}{a+c}},\sqrt{\frac{c}{a+b}} \right )=(x,y,z)\)
\(\Rightarrow \left\{\begin{matrix} x^2=\frac{a}{b+c}\\ y^2=\frac{b}{a+c}\\ z^2=\frac{c}{a+b}\end{matrix}\right.\Rightarrow \frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}=2\)
\(\Leftrightarrow (1-\frac{1}{x^2+1})+(1-\frac{1}{y^2+1})+(1-\frac{1}{z^2+1})=1\)
\(\Leftrightarrow \frac{x^2}{x^2+1}+\frac{y^2}{y^2+1}+\frac{z^2}{z^2+1}=1\)
BĐT cần chứng minh tương đương:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 2(x+y+z)(\star)\)
Áp dụng BĐT Bunhiacopxky:
\(\left ( \frac{x^2}{x^2+1}+\frac{y^2}{y^2+1}+\frac{z^2}{z^2+1} \right )(x^2+1+y^2+1+z^2+1)\geq (x+y+z)^2\)
\(\Leftrightarrow x^2+1+y^2+1+z^2+1\geq (x+y+z)^2\)
\(\Leftrightarrow xy+yz+xz\leq \frac{3}{2}\)
Kết hợp với hệ quả của BĐT AM-GM :
\((xy+yz+xz)^2\geq 3xyz(x+y+z)\)
\(\Rightarrow xy+yz+xz\geq \frac{3xyz(x+y+z)}{xy+yz+xz}\geq \frac{3xyz(x+y+z)}{\frac{3}2{}}=2xyz(x+y+z)\)
\(\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{2xyz(x+y+z)}{xyz}=2(x+y+z)\)
Do đó BĐT \((\star)\) được chứng minh.
Bài toán hoàn thành. Dấu bằng xảy ra khi \(a=b=c\)
Cho a,b,c > 0 thỏa mãn \(a\sqrt{\dfrac{b}{c}}+b\sqrt{\dfrac{c}{a}}+c\sqrt{\dfrac{a}{b}}=3\). Chứng minh rằng:
\(N=\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}+\dfrac{c^4}{a^2}\ge3\)
Áp dụng \(x^2+y^2+z^2\ge xy+yz+zx\) và \(x^2+y^2+z^2\ge\dfrac{1}{3}\left(x+y+z\right)^2\)
\(N\ge\dfrac{a^2b}{c}+\dfrac{b^2c}{a}+\dfrac{c^2a}{b}\ge\dfrac{1}{3}\left(a\sqrt{\dfrac{b}{c}}+b\sqrt{\dfrac{c}{a}}+c\sqrt{\dfrac{a}{b}}\right)^2=3\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho \(a,b>0\); \(c< 0\). Chứng minh rằng:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\)
Cần cm:
\(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\\ \Leftrightarrow a+b=a+b+2c+2\sqrt{\left(a+c\right)\left(b+c\right)}\\ \Leftrightarrow2c+2\sqrt{ab+ac+bc+c^2}=0\\ \Leftrightarrow2c+2\sqrt{c^2}=0\\ \Leftrightarrow2c+2\left|c\right|=0\\ \Leftrightarrow2c-2c=0\left(c< 0\right)\\ \Leftrightarrow0=0\left(luôn.đúng\right)\)
Vậy đẳng thức đc cm
Cho a, b, c > 0. Chứng minh rằng: \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Ta co : √a/√(b+c)=a/√a.(b+c)=2a/2√a.(b+c)≥2a/(a+b+c)
Vi a,b,c>0 nen √a/√(b+c)>2a/(a+b+c)
Tuong tu √b/√(b+c)>2b(a+b+c)
√c/√(a+b)>2c/(a+b+c)
=> VT>2a/(a+b+c) + 2b/(a+b+c) + 2c/(a+b+c)=2.(a+b+c)/(a+b+c)=2
Cho a,b,c >0 Chứng minh rằng:
a) \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b+c}{\sqrt[3]{abc}}\)
b) \(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}\ge\sqrt{3\left(a^2+b^2+c^2\right)}\)
Cho a,b,c > 0 và ab + bc + ac = 1. Chứng minh rằng :\(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{3}{2}\)
\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+ac+bc}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)=\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\) Chứng minh tương tự ta được:
\(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{b+a}+\dfrac{b}{b+c}\right);\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+a}+\dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)=\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\left(1+1+1\right)=\dfrac{3}{2}\) Dấu = xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{\sqrt{3}}\)
\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
Tương tự: \(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\) ; \(\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)
Cộng vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Cho a,b,c>0 thỏa mãn: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1980\)
Chứng minh rằng: \(\dfrac{\sqrt{b^2+2a^2}}{ab}+\dfrac{\sqrt{c^2+2b^2}}{bc}+\dfrac{\sqrt{a^2+2c^2}}{ac}\ge1980\sqrt{3}\)
\(\dfrac{\sqrt{b^2+a^2+a^2}}{ab}\ge\dfrac{\sqrt{\dfrac{1}{3}\left(b+a+a\right)^2}}{ab}=\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)\)
Tương tự: \(\dfrac{\sqrt{c^2+2b^2}}{bc}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)\) ; \(\dfrac{\sqrt{a^2+2c^2}}{ac}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)\)
Cộng vế với vế:
\(VT\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1980\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{3}{1980}\)
Cho a,b,c là các số thực dương thỏa mãn \(\sqrt{a}+\sqrt{b}+\sqrt{c}=2\). Chứng minh rằng:\(\dfrac{a+b}{\sqrt{a}+\sqrt{b}}+\dfrac{b+c}{\sqrt{b}+\sqrt{c}}+\dfrac{c+a}{\sqrt{c}+\sqrt{a}}\le4\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{b}}+\dfrac{\left(\sqrt{b}-1\right)^2}{\sqrt{c}}+\dfrac{\left(\sqrt{c}-1\right)^2}{\sqrt{a}}\right)\)