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Áp dụng bất đẳng thức Cauchy-Schwarz:
\(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}=\dfrac{1}{a}+\dfrac{4}{2b}+\dfrac{9}{3c}\ge\dfrac{\left(1+2+3\right)^2}{a+2b+3c}=\dfrac{36}{a+2b+3c}\)
\(\dfrac{2}{a}+\dfrac{3}{b}+\dfrac{1}{c}=\dfrac{4}{2a}+\dfrac{9}{3b}+\dfrac{1}{c}\ge\dfrac{\left(2+3+1\right)^2}{2a+3b+c}=\dfrac{36}{2a+3b+c}\)
\(\dfrac{3}{a}+\dfrac{1}{b}+\dfrac{2}{c}=\dfrac{9}{3a}+\dfrac{1}{b}+\dfrac{4}{2c}\ge\dfrac{\left(3+1+2\right)^2}{3a+b+2c}=\dfrac{36}{3a+2b+c}\)
Cộng theo vế: \(6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge36F\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge6F\)
Mặt khác: \(ab+bc+ac=3abc\Leftrightarrow\dfrac{ab+bc+ac}{abc}=3\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
\(\Rightarrow18\ge36F\Leftrightarrow F\le\dfrac{1}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
b) \(\dfrac{1}{3a+2b+c}\le\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{1}{36}\left(\dfrac{3}{a}+\dfrac{2}{b}+\dfrac{1}{c}\right)\)
Tương tự cho 2 cái kia rồi cộng lại
\(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}.16=\dfrac{8}{3}\)
Đẳng thức xảy ra \(\Leftrightarrow\) ... \(\Leftrightarrow a=b=c=\dfrac{3}{16}\)
Dean thật, gõ gần xong rồi tự nhiên nó tạch, phải gõ lại -.-
Từ gt, ta suy ra:
\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right].\dfrac{1}{2}=0\)(Tự phân tích, không còn kiên nhẫn để gõ lại)
Mà a+b+c khác 0 => a=b=c
Thay vào thì C=8
bai 2 :
dat cac tich ab , bc , ca lan luot la x,y,z ( khac 0 )
thay vao ta dc : x^3+y^3+z^3=3xyz
=> (x+y)(x^2-2xy+y^2)+z^3-3xyz=0
=>(x+y)(x^2+2xy+y^2)+z^3-3xy(x+y)-3xyz=0
=》（x+y+z)【（x+y）^2 -(x+y)z+z^2】-3xy(x+y+z)=0
=>(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0
=>\(\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]\)=0
=> x+y+z=0 hoac x=y=z
TH1 : a+b+c=0
=>P=-1
TH2 : a=b=c
=>P=8
12. Ta có \(ab\le\frac{a^2+b^2}{2}\)
=> \(a^2-ab+3b^2+1\ge\frac{a^2}{2}+\frac{5}{2}b^2+1\)
Lại có \(\left(\frac{a^2}{2}+\frac{5}{2}b^2+1\right)\left(\frac{1}{2}+\frac{5}{2}+1\right)\ge\left(\frac{a}{2}+\frac{5}{2}b+1\right)^2\)
=> \(\sqrt{a^2-ab+3b^2+1}\ge\frac{a}{4}+\frac{5b}{4}+\frac{1}{2}\)
=> \(\frac{1}{\sqrt{a^2-ab+3b^2+1}}\le\frac{4}{a+b+b+b+b+b+1+1}\le\frac{4}{64}.\left(\frac{1}{a}+\frac{5}{b}+2\right)\)
Khi đó
\(P\le\frac{1}{16}\left(6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+6\right)\le\frac{3}{2}\)
Dấu bằng xảy ra khi a=b=c=1
Vậy \(MaxP=\frac{3}{2}\)khi a=b=c=1
13. Ta có \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le1\)
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{9}{a+b+c+3}\)( BĐT cosi)
=> \(1\ge\frac{9}{a+b+c+3}\)
=> \(a+b+c\ge6\)
Ta có \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
=> \(\frac{a^3-b^3}{a^2+ab+b^2}=a-b\)
Tương tự \(\frac{b^3-c^3}{b^2+bc+c^2}=b-c\),,\(\frac{c^3-a^2}{c^2+ac+a^2}=c-a\)
Cộng 3 BT trên ta có
\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+c^2}=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{c^2+bc+b^2}+\frac{a^3}{a^2+ac+c^2}\)
Khi đó \(2P=\frac{a^3+b^3}{a^2+ab+b^2}+...\)
=> \(2P=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}+....\)
Xét \(\frac{a^2-ab+b^2}{a^2+ab+b^2}\ge\frac{1}{3}\)
<=> \(3\left(a^2-ab+b^2\right)\ge a^2+ab+b^2\)
<=> \(a^2+b^2\ge2ab\)(luôn đúng )
=> \(2P\ge\frac{1}{3}\left(a+b+b+c+a+c\right)=\frac{2}{3}.\left(a+b+c\right)\ge4\)
=> \(P\ge2\)
Vậy \(MinP=2\)khi a=b=c=2
Lưu ý : Chỗ .... là tương tự
+ \(2a+b+c=\left(a+b\right)+\left(a+c\right)\)
\(\ge2\sqrt{\left(a+b\right)\left(a+c\right)}\) ( theo AM-GM )
\(\Rightarrow\left(2a+b+c\right)^2\ge4\left(a+b\right)\left(a+c\right)\)
\(\Rightarrow\frac{1}{\left(2a+b+c\right)^2}\le\frac{1}{4\left(a+b\right)\left(a+c\right)}\)
Dấu "=" xảy ra \(\Leftrightarrow b=c\)
+ Tương tự : \(\frac{1}{\left(2b+c+a\right)^2}\le\frac{1}{4\left(a+b\right)\left(b+c\right)}\). Dấu "=" xảy ra <=> a = c
\(\frac{1}{\left(2c+a+b\right)^2}\le\frac{1}{4\left(a+c\right)\left(b+c\right)}\). Dấu "=" xảy ra \(\Leftrightarrow a=b\)
Do đó : \(P\le\frac{1}{4}\left(\frac{1}{\left(a+b\right)\left(a+c\right)}+\frac{1}{\left(a+b\right)\left(b+c\right)}+\frac{1}{\left(a+c\right)\left(b+c\right)}\right)\)
\(\Rightarrow P\le\frac{1}{2}\cdot\frac{a+b+c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}\)\(=8abc\)
\(\Rightarrow P\le\frac{a+b+c}{16abc}\)
+ \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\). Dấu :=" xảy ra \(\Leftrightarrow a=b\)
\(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\). Dấu "=" xảy ra <=> b = c
\(\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ca}\). Dấu "=" xảy ra <=> c = a
\(\Rightarrow2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\Rightarrow3\ge\frac{a+b+c}{abc}\) \(\Rightarrow a+b+c\le3abc\)
\(\Rightarrow P\le\frac{3abc}{16abc}=\frac{3}{16}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Sửa đề: Cho thêm a,b,c dương
Áp dụng BĐT AM-GM ta có:
\(a^2+2b^2+3c^2\ge6\sqrt[6]{a^2\cdot b^2\cdot b^2\cdot c^2\cdot c^2\cdot c^2}=6\sqrt[6]{a^2b^4c^6}\)
\(\Rightarrow3abc\ge6\sqrt[6]{a^2b^4c^6}\Leftrightarrow abc\ge2\sqrt[6]{a^2b^4c^6}\)
\(\Leftrightarrow a^6b^6c^6\ge64a^2b^4c^6\Leftrightarrow a^4b^2\ge64\Leftrightarrow a^2b\ge8\)
\(\Rightarrow2\le\sqrt[3]{a\cdot a\cdot b}\le\dfrac{2a+b}{3}\Leftrightarrow2a+b\ge6\)
Khi đó ta có: \(P=2a+\dfrac{8}{a}+\dfrac{3b}{2}+\dfrac{6}{b}+c+\dfrac{4}{c}+\dfrac{2a+b}{2}\)
Áp dụng tiếp BĐT AM-GM ta có:
\(P\ge2\sqrt{2a\cdot\dfrac{8}{a}}+2\sqrt{\dfrac{3b}{2}\cdot\dfrac{6}{b}}+2\sqrt{c\cdot\dfrac{4}{c}}+\dfrac{6}{2}\left(2a+b\ge6\right)\)
\(=2\sqrt{16}+2\sqrt{9}+2\sqrt{4}+3=8+6+4+3=21\)
Đẳng thức xảy ra khi \(a=b=c=2\)
Người ta bảo tính giá trị của biểu thức chứ có bảo tìm cực trị của nó đâu.