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cho số dương a,b,c. Tìm GTLN : \(\dfrac{ab}{a^2+ab+bc}+\dfrac{bc}{b^2+bc+ca}+\dfrac{ca}{c^2+ca+ab}\)
\(VT=\dfrac{1}{\dfrac{a}{b}+\dfrac{c}{a}+1}+\dfrac{1}{\dfrac{b}{c}+\dfrac{a}{b}+1}+\dfrac{1}{\dfrac{c}{a}+\dfrac{b}{c}+1}\)
\(\left(\dfrac{a}{b},\dfrac{b}{c},\dfrac{c}{a}\right)\rightarrow\left(x^3,y^3,z^3\right)\)\(\Rightarrow xyz=1\).
\(VT=\sum\dfrac{1}{x^3+y^3+1}\le\sum\dfrac{1}{xy\left(x+y\right)+xyz}=\sum\dfrac{z}{x+y+z}=1\)
Dấu = xảy ra khi x=y=z=1 hay a=b=c
\(P=\sqrt{\dfrac{ab}{c+ab}}+\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ca}{b+ca}}\)
\(=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}+\sqrt{\dfrac{bc}{a\left(a+b+c\right)+bc}}+\sqrt{\dfrac{ca}{b\left(a+b+c\right)+ca}}\)
\(=\sqrt{\dfrac{ab}{\left(b+c\right)\left(c+a\right)}}+\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{ca}{\left(a+b\right)\left(b+c\right)}}\)
\(\le\dfrac{1}{2}\left(\dfrac{b}{b+c}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{a+b}+\dfrac{c}{b+c}+\dfrac{a}{a+b}\right)=\dfrac{1}{2}\)
\("=" \Leftrightarrow a=b=c=\frac{1}{3}\)
Áp dụng BĐT AM-Gm: ( dạng \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\))
\(VT=\sum\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{1}{9}\left(\sum\dfrac{a}{2}+\sum\left[\dfrac{ab}{a+c}+\dfrac{bc}{a+c}\right]\right)\)
\(=\dfrac{1}{9}\left(\dfrac{a+b+c}{2}+a+b+c\right)=\dfrac{1}{6}\left(a+b+c\right)\)
\(\le\dfrac{1}{6}\sqrt{3\left(a^2+b^2+c^2\right)}=1\) (đpcm)
Dấu = xảy ra khi a=b=c=2
Lời giải:
Ta có:
\(P=\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}=\frac{(ab)^2+(bc)^2+(ca)^2}{abc}\)
Xét tử số:
\(\text{TS}=(ab)^2+(bc)^2+(ca)^2\)
\(\Rightarrow \text{TS}^2=a^4b^4+b^4c^4+c^4a^4+2(a^2b^4c^2+a^2b^2c^4+a^4b^2c^2)\)
Áp dụng BĐT AM-GM ta có:
\(\left\{\begin{matrix} a^4b^4+b^4c^4\geq 2a^2b^4c^2\\ b^4c^4+c^4a^4\geq 2a^2b^2c^4\\ c^4a^4+a^4b^4\geq 2a^4b^2c^2\end{matrix}\right.\)
Cộng theo vế và rút gọn:
\(\Rightarrow a^4b^4+b^4c^4+c^4a^4\geq a^2b^4c^2+a^2b^2c^4+a^4b^2c^2\)
Do đó:
\(\text{TS}^2\geq 3(a^2b^4c^2+a^2b^2c^4+a^4b^2c^2)=3a^2b^2c^2(a^2+b^2+c^2)=3a^2b^2c^2\)
\(\Rightarrow \text{TS}\geq \sqrt{3}abc\)
\(\Rightarrow P\geq \sqrt{3}\)
Vậy \(P_{\min}=\sqrt{3}\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)
Cách khác:
\(P^2=\dfrac{a^2b^2}{c^2}+\dfrac{b^2c^2}{a^2}+\dfrac{c^2a^2}{b^2}+2\left(a^2+b^2+c^2\right)\)
Áp dụng BĐT Cauchy:
\(\dfrac{a^2b^2}{c^2}+\dfrac{b^2c^2}{a^2}\ge2b^2\)
CMTT\(\Rightarrow\)\(\dfrac{a^2b^2}{c^2}+\dfrac{b^2c^2}{a^2}+\dfrac{a^2c^2}{b^2}\ge a^2+b^2+c^2\)
\(\Rightarrow P^2\ge3\Rightarrow P\ge\sqrt{3}\)
Dấu"=" xảy ra\(\Leftrightarrow\)a=b=c=\(\dfrac{1}{\sqrt{3}}\)
Ta có:
\(\dfrac{a}{bc}+\dfrac{b}{ca}\ge2\sqrt{\dfrac{ab}{abc^2}}=\dfrac{2}{c}\)
Tương tự: \(\dfrac{a}{bc}+\dfrac{c}{ab}\ge\dfrac{2}{b}\) ; \(\dfrac{b}{ca}+\dfrac{c}{ab}\ge\dfrac{2}{a}\)
Cộng vế với vế: \(\Rightarrow\dfrac{a}{bc}+\dfrac{b}{ca}+\dfrac{c}{ab}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow P\ge\dfrac{a^2+b^2+c^2}{2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(a^2+\dfrac{1}{a}+\dfrac{1}{a}\right)+\dfrac{1}{2}\left(a^2+\dfrac{1}{b}+\dfrac{1}{b}\right)+\dfrac{1}{2}\left(c^2+\dfrac{1}{c}+\dfrac{1}{c}\right)\)
\(\Rightarrow P\ge\dfrac{1}{2}.3\sqrt[3]{\dfrac{a^2}{a^2}}+\dfrac{1}{2}.3\sqrt[3]{\dfrac{b^2}{b^2}}+\dfrac{1}{2}.3\sqrt[3]{\dfrac{c^2}{c^2}}=\dfrac{9}{2}\)
\(P_{min}=\dfrac{9}{2}\) khi \(a=b=c=1\)
Bài 1
\(VT=\dfrac{a^2}{ab^2+abc+ac^2}+\dfrac{b^2}{c^2b+abc+a^2b}+\dfrac{c^2}{a^2c+abc+b^2c}\)
Áp dụng bđt Cauchy dạng phân thức
\(\Rightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{ab\left(a+b\right)+abc+ac\left(a+c\right)+abc+bc\left(b+c\right)+abc}\)
\(\Leftrightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}\)
\(\Leftrightarrow VT\ge\dfrac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)
Dấu ''='' xảy ra khi \(a=b=c\)
Bài 2
\(VT=\left(\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}\right)\left[\left(\dfrac{\sqrt{a}}{b+c}\right)^2+\left(\dfrac{\sqrt{b}}{c+a}\right)^2+\left(\dfrac{\sqrt{c}}{a+b}\right)^2\right]\)
Áp dụng bđt Bunhiacopxki ta có
\(VT\ge\left(\sqrt{a}.\dfrac{\sqrt{a}}{b+c}+\sqrt{b}.\dfrac{\sqrt{b}}{c+a}+\sqrt{c}.\dfrac{\sqrt{c}}{a+b}\right)^2\)
\(\Leftrightarrow VT\ge\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\)
Xét \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
Áp dụng bđt Cauchy dạng phân thức ta có
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ac\right)}=\dfrac{3}{2}\)
\(\Rightarrow\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\ge\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)
\(\Rightarrow VT\ge\dfrac{9}{4}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c\)
Link hình: file:///C:/Users/THAOCAT/Pictures/Screenshots/Screenshot%20(1231).png
Từ O kẻ \(OD\perp BC,OE\perp AC,OF\perp AB\left(D\in BC,E\in AC,F\in AB\right)\)
Lấy các điểm D', E', F' lần lượt đối xứng với O qua BC, AC, AB
\(\Delta AFO\)và \(\Delta AEO\)vuông có AO là phân giác nên \(\Delta AFO=\Delta AEO\)từ đó suy ra được: \(\Delta AFO=\Delta AEO=\Delta AFF'=\Delta AEE'\)
\(\Delta ABC\)và \(\Delta OAE'\)có \(\widehat{BAC}=\widehat{OAE'}\)nên \(\frac{S_{OAE'}}{S_{ABC}}=\frac{AO.AE'}{AB.AC}=\frac{OA^2}{bc}\)hay \(\frac{S_{AFOE}}{S_{ABC}}=\frac{OA^2}{bc}\)
Tương tự: \(\frac{S_{BFOD}}{S_{ABC}}=\frac{OB^2}{ca}\); \(\frac{S_{CEOD}}{S_{ABC}}=\frac{OC^2}{ab}\)
Từ đó suy ra \(K=1\)