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Ta có:
\(\left(a^2+b+c\right)\left(1+b+c\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+b+c}}\le\dfrac{a\sqrt{1+b+c}}{a+b+c}\)
Tương tự: \(\dfrac{b}{\sqrt{b^2+a+c}}\le\dfrac{b\sqrt{1+c+a}}{a+b+c}\) ; \(\dfrac{c}{\sqrt{c^2+b+a}}\le\dfrac{c\sqrt{1+a+b}}{a+b+c}\)
Cộng vế:
\(P\le\dfrac{a\sqrt{1+b+c}+b\sqrt{1+c+a}+c\sqrt{1+a+b}}{a+b+c}\)
Lại có:
\(a\sqrt{1+b+c}+b\sqrt{1+c+a}+c\sqrt{1+a+b}\)
\(=\sqrt{a}.\sqrt{a+ab+ac}+\sqrt{b}.\sqrt{b+bc+ab}+\sqrt{c}.\sqrt{c+ac+bc}\)
\(\le\sqrt{\left(a+b+c\right)\left(a+b+c+2ab+2bc+2ca\right)}\)
\(\Rightarrow P\le\dfrac{\sqrt{\left(a+b+c\right)\left(a+b+c+2ab+bc+ca\right)}}{a+b+c}=\sqrt{\dfrac{a+b+c+2ab+2bc+2ca}{a+b+c}}\)
Do đó ta chỉ cần chứng minh:
\(\dfrac{a+b+c+2ab+2bc+2ca}{a+b+c}\le3\Leftrightarrow a+b+c\ge ab+bc+ca\)
Thật vậy:
\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)=\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)
\(\Rightarrow a+b+c\ge ab+bc+ca\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
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 \(VT>2\)
Bất đẳng thức cần chứng minh tương đương với:
\(\dfrac{a}{a+3\sqrt{bc}}+\dfrac{b}{b+3\sqrt{ca}}+\dfrac{c}{c+3\sqrt{ab}}\)
Ta áp dụng bất đẳng thức Cô si dạng \(2\sqrt{xy}\le x+y\) cho các căn thức ở mẫu, khi đó ta được:
\(\dfrac{a}{a+3\sqrt{bc}}+\dfrac{b}{b+3\sqrt{ca}}+\dfrac{c}{c+3\sqrt{ab}}\ge\) với biểu thức
\(\dfrac{2a}{2a+3b+3c}+\dfrac{2b}{3a+2b+3c}+\dfrac{2c}{3a+3b+2c}\)
Khi đó ta cần chứng minh:
\(\dfrac{2a}{2a+3b+3c}+\dfrac{2b}{3a+2b+3c}+\dfrac{2c}{3a+3b+2c}\ge\dfrac{3}{4}\)
Đặt: \(\left\{{}\begin{matrix}x=2a+3b+3c\\y=3a+2b+3c\\z=3a+3b+2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a=\dfrac{1}{4}\left(3y+3z-5x\right)\\2b=\dfrac{1}{4}\left(3z+3x-5y\right)\\2c=\dfrac{1}{4}\left(3x+3y-5z\right)\end{matrix}\right.\)
Khi đó đẳng thức trên được viết lại thành:
\(\dfrac{3y+3z-5x}{4x}+\dfrac{3z+3x-5y}{4y}+\dfrac{3x+3y-5z}{4z}\ge\dfrac{3}{4}\)
Hay: \(3\left(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{x}{z}+\dfrac{z}{x}\right)-15\ge3\)
Bất đẳng thức cuối cùng luôn đúng theo bất đẳng thức Cô si.
Vậy bất đẳng thức được chứng minh. Đẳng thức xảy ra khi và chỉ khi \(a=b=c\)
Đặt \(x=\sqrt{a};y=\sqrt{b};z=\sqrt{c}\)
Khi đó bđt đã tro chở thành:
\(\dfrac{yz}{x^2+3yz}+\dfrac{zx}{y^2+3zx}+\dfrac{xy}{z^2+3xy}\le\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{yz}{x^2+3yz}+\dfrac{1}{3}-\dfrac{zx}{y^2+3zx}+\dfrac{1}{3}-\dfrac{xy}{z^2+3xy}\ge1-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{x^2}{x^2+3yz}+\dfrac{y^2}{y^2+3zx}+\dfrac{z^2}{z^2+3xy}\ge\dfrac{3}{4}\) (đpcm)
\(\dfrac{a}{a+2\sqrt{\left(a+bc\right)}}=\dfrac{a}{a+2\sqrt{a\left(a+b+c\right)+bc}}=\dfrac{a}{a+2\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(=\dfrac{a}{a+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\)
\(\le\dfrac{a}{5^2}\left(\dfrac{1}{a}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\right)\)
\(=\dfrac{a}{25}\left(\dfrac{1}{a}+\dfrac{8}{\sqrt{\left(a+b\right)\left(a+c\right)}}\right)=\dfrac{1}{25}+\dfrac{8}{25}.\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
Tương tự:
\(\dfrac{b}{b+2\sqrt{b+ac}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\)
\(\dfrac{c}{c+2\sqrt{c+ab}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)
Cộng vế:
\(P\le\dfrac{3}{25}+\dfrac{4}{25}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{15}{25}=\dfrac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Lời giải:
Do $a+b+c=1$ nên:
\(\text{VT}=\sqrt{\frac{ab}{c(a+b+c)+ab}}+\sqrt{\frac{bc}{a(a+b+c)+bc}}+\sqrt{\frac{ca}{b(a+b+c)+ac}}\)
\(=\sqrt{\frac{ab}{(c+a)(c+b)}}+\sqrt{\frac{bc}{(a+b)(a+c)}}+\sqrt{\frac{ca}{(b+c)(b+a)}}\)
Áp dụng BĐT AM-GM:
\(\sqrt{\frac{ab}{(c+a)(c+b)}}\leq \frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)
\(\sqrt{\frac{bc}{(a+b)(a+c)}}\leq \frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{c+a}\right)\)
\(\sqrt{\frac{ca}{(b+c)(b+a)}}\leq \frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{b+a}\right)\)
Cộng theo vế:
\(\Rightarrow \text{VT}\leq \frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
Áp dụng bất đẳng thức Bunyakovsky
\(\Rightarrow\sqrt{\left(\dfrac{8}{a^2}+\dfrac{9b^2}{2}+\dfrac{c^2a^2}{4}\right)\left[\left(\sqrt{2}\right)^2+\left(3\sqrt{2}\right)^2+2^2\right]}\ge\left(\sqrt{\dfrac{4}{a}+9b+ca}\right)^2\)
\(\Leftrightarrow2\sqrt{6}\sqrt{\dfrac{8}{a^2}+\dfrac{9b^2}{2}+\dfrac{c^2a^2}{4}}\ge\dfrac{4}{a}+9b+ac\)
Tương tự ta có \(\left\{{}\begin{matrix}2\sqrt{6}\sqrt{\left(\dfrac{8}{b^2}+\dfrac{9c^2}{2}+\dfrac{a^2b^2}{4}\right)}\ge\dfrac{4}{b}+9c+ab\\2\sqrt{6}\sqrt{\left(\dfrac{8}{c^2}+\dfrac{9a^2}{2}+\dfrac{b^2c^2}{4}\right)}\ge\dfrac{4}{c}+9a+bc\end{matrix}\right.\)
\(\Rightarrow2\sqrt{6}S\ge\dfrac{4}{a}+9a+\dfrac{4}{b}+9b+\dfrac{4}{c}+9c+ab+bc+ac\)
\(\Leftrightarrow2\sqrt{6}S\ge\dfrac{4}{a}+a+8a+\dfrac{4}{b}+b+8b+\dfrac{4}{c}+c+8c+ab+bc+ca\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{a}+a\ge2\sqrt{4}=4\\\dfrac{4}{b}+b\ge2\sqrt{4}=4\\\dfrac{4}{c}+c\ge2\sqrt{4}=4\end{matrix}\right.\)
\(\Rightarrow\dfrac{4}{a}+a+8a+\dfrac{4}{b}+b+8b+\dfrac{4}{c}+c+8c+ab+bc+ca\ge12+8a+8b+8c+ab+bc+ac\)
\(\Rightarrow2\sqrt{6}S\ge12+8a+8b+8c+ab+bc+ac\)
\(\Leftrightarrow2\sqrt{6}S\ge12+2a+bc+2b+ac+2c+ab+6\left(a+b+c\right)\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow2a+bc\ge2\sqrt{2abc}\)
Tượng tự ta có \(2b+ac\ge2\sqrt{2abc}\) ; \(2c+ab\ge2\sqrt{2abc}\)
\(\Rightarrow12+2a+bc+2b+ac+2c+ab+6\left(a+b+c\right)\ge6\left(a+b+c+\sqrt{2abc}\right)+12\)
\(\Rightarrow2\sqrt{6}S\ge6\left(a+b+c+\sqrt{2abc}\right)+12\)
Theo đề bài ta có \(a+b+c+\sqrt{2abc}\ge10\)
\(\Rightarrow6\left(a+b+c+\sqrt{2abc}\right)+12\ge72\)
\(\Rightarrow S\ge\dfrac{72}{2\sqrt{6}}=6\sqrt{6}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=2\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a^3}{\sqrt{b^2+3}}+\dfrac{a^3}{\sqrt{b^2+3}}+\dfrac{b^2+3}{8}\ge\dfrac{3a^2}{2}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b^3}{\sqrt{c^2+3}}+\dfrac{b^3}{\sqrt{c^2+3}}+\dfrac{c^2+3}{8}\ge\dfrac{3b^2}{2};\dfrac{c^3}{\sqrt{a^2+3}}+\dfrac{c^3}{\sqrt{a^2+3}}+\dfrac{a^2+3}{8}\ge\dfrac{3c^2}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(2P+\dfrac{a^2+b^2+c^2+9}{8}\ge\dfrac{3\left(a^2+b^2+c^2\right)}{2}\)
\(\Leftrightarrow P\ge\dfrac{\dfrac{3\left(a^2+b^2+c^2\right)}{2}-\dfrac{a^2+b^2+c^2+9}{8}}{2}=\dfrac{3}{2}\)
@DƯƠNG PHAN KHÁNH DƯƠNG
\(a;b;c\ge0\)thỏa mãn \(ab+bc+ca=1\). CMR \(\dfrac{1}{2a+2bc+1}+\dfrac{1}{2b+2ca+1}+\dfrac{1}{2c+2ab+1}\ge1\)
Đảm bảo an ninh :))
a) Ta có:
\(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^2}{9}\ge\dfrac{\left(ab+bc+ca\right)}{3}\)
\(\Leftrightarrow\dfrac{a+b+c}{3}\ge\sqrt{\dfrac{ab+bc+ca}{3}}\)
Đẳng thức xảy ra khi $a=b=c.$
b) BĐT \(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
Hay là \(2\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\),
đúng.
Đẳng thức xảy ra khi $a=b=c.$
c) \(\Leftrightarrow\dfrac{\left(x^2+2\right)^2}{x^2+1}\ge4\Leftrightarrow x^4+4x^2+4\ge4x^2+4\Leftrightarrow x^4\ge0\)
Đẳng thức xảy ra khi $x=0.$
d) Xét hiệu hai vế đi bạn.
Ta có :
\(\sqrt{\dfrac{c}{a+b}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{a}{b+c}}=\dfrac{c}{\sqrt{c\left(a+b\right)}}+\dfrac{b}{\sqrt{b\left(c+a\right)}}+\dfrac{a}{\sqrt{a\left(b+c\right)}}\)Áp dụng BĐT Cauchy :
\(\Rightarrow\dfrac{c}{\sqrt{c\left(a+b\right)}}+\dfrac{b}{\sqrt{b\left(c+a\right)}}+\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2c}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2a}{a+b+c}=2\)Đấu đẳng thức xảy ra khi \(\left\{{}\begin{matrix}a=b+c\\b=c+a\\c=a+b\end{matrix}\right.\)\(\Rightarrow a+b+c=2\left(a+b+c\right)\Rightarrow1=2\) Vậy dấu đẳng thức không xảy ra
Ta phải chứng minh :
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)
ta chứng minh bất đẳng thức phụ sau :
do \(\dfrac{a}{a+b}< 1\Rightarrow\dfrac{a}{a+b}< \dfrac{a+c}{a+b+c}\)
tương tự : \(\dfrac{b}{b+c}< \dfrac{b+a}{a+b+c}\); \(\dfrac{c}{c+a}< \dfrac{c+b}{a+b+c}\)
cộng ba vế BĐT lại ta có đpcm
thaks bn nhìu nha