Có \(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\ge9\)

\(\Leftrightarrow\dfrac{a+1}{a}.\dfrac{b+1}{b}\ge9\)

\(\Leftrightarrow ab+a+b+1\ge9ab\) ( vì ab >0)

\(\Leftrightarrow a+b+1\ge8ab\)

\(\Leftrightarrow2\ge8ab\) \(\left(a+b=1\right)\)

\(\Leftrightarrow1\ge4ab\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\) \(\left(a+b=1\right)\)

\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\) ( luôn đúng)

\(\Leftrightarrowđpcm\)

\(VT=\dfrac{a^2}{a+abc}+\dfrac{b^2}{b+abc}+\dfrac{c^2}{c+abc}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+3abc}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+\dfrac{1}{9}\left(a+b+c\right)^3}=\dfrac{1^2}{1+\dfrac{1}{9}.1^3}=\dfrac{9}{10}\)

=9/10

Với a;b;c dương:

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(=\left(a+b+c\right)\left(ab+bc+ca\right)-\sqrt[3]{abc}.\sqrt[3]{ab.bc.ca}\)

\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\dfrac{1}{3}\left(a+b+c\right).\dfrac{1}{3}\left(ab+bc+ca\right)\)

\(=\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)

Đặt vế trái BĐT là P, ta có:

\(\dfrac{ab}{1-c^2}=\dfrac{ab}{\left(1-c\right)\left(1+c\right)}=\dfrac{ab}{\left(a+b\right)\left(a+c+b+c\right)}=\dfrac{ab}{\sqrt{a+b}.\sqrt{a+b}\left(a+c+b+c\right)}\)

\(\le\dfrac{ab}{\sqrt[]{2\sqrt[]{ab}}.\sqrt[]{a+b}.2\sqrt[]{\left(a+c\right)\left(b+c\right)}}=\dfrac{\sqrt[4]{\left(ab\right)^3}}{2\sqrt[]{2}.\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

Tương tự:

\(\dfrac{bc}{1-a^2}\le\dfrac{\sqrt[4]{\left(bc\right)^3}}{2\sqrt[]{2}.\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

\(\dfrac{ca}{1-b^2}\le\dfrac{\sqrt[4]{\left(ca\right)^3}}{2\sqrt[]{2}.\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

Cộng vế:

\(P\le\dfrac{\sqrt[4]{\left(ab\right)^3}+\sqrt[4]{\left(bc\right)^3}+\sqrt[4]{\left(ca\right)^3}}{2\sqrt[]{2}.\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

Nên ta chỉ cần chứng minh:

\(\sqrt[4]{\left(ab\right)^3}+\sqrt[4]{\left(bc\right)^3}+\sqrt[4]{\left(ca\right)^3}\le\dfrac{3}{2\sqrt[]{2}}\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(\Leftrightarrow\left(\sqrt[4]{\left(ab\right)^3}+\sqrt[4]{\left(bc\right)^3}+\sqrt[4]{\left(ca\right)^3}\right)^2\le\dfrac{9}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Mà \(\dfrac{9}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)\)

Nên ta chỉ cần chứng minh:

\(\left(\sqrt[4]{\left(ab\right)^3}+\sqrt[4]{\left(bc\right)^3}+\sqrt[4]{\left(ca\right)^3}\right)^2\le\left(a+b+c\right)\left(ab+bc+ca\right)\)

Thật vậy:

\(\left(\sqrt[4]{ab}.\sqrt[]{ab}+\sqrt[4]{bc}.\sqrt[]{bc}+\sqrt[4]{ca}.\sqrt[]{ca}\right)^2\le\left(\sqrt[]{ab}+\sqrt[]{bc}+\sqrt[]{ca}\right)\left(ab+bc+ca\right)\)

\(\le\left(a+b+c\right)\left(ab+bc+ca\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

dễ làm

1:5/6va 1/8

2:55 va 99

3:3 va 7

mình làm rồi bạn ạ,mình mới học sag ny, cho minh nha

trình bày ra chứ

phải trình bày

Mình bổ sung một cách làm khác nhé.

Áp dụng BĐT Cô-si cho 3 số dương \(a,b,c\), ta có \(a+b+c\ge3\sqrt[3]{abc}\) \(\Rightarrow1\ge3\sqrt[3]{abc}\)      (1)

Áp dụng BĐT Cô-si cho 3 số dương \(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\) ta có \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)           (2)

Nhân theo vế của các BĐT (1) và (2), ta được \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\) (đpcm)

Đẳng thức xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{3}\)

\(Ta\) có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)

\(=1+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{a}{b}+\dfrac{c}{b}+1+\dfrac{a}{c}+\dfrac{b}{c}+1\)

\(=\left(1+1+1\right)+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)+\left(\dfrac{c}{b}+\dfrac{b}{c}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\)

\(Ta\) có : \(\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge2\Leftrightarrow\dfrac{a^2+b^2}{ab}-2\ge0\Leftrightarrow\dfrac{a^2-2ab+b^2}{ab}\ge0\)

\(cmt\) \(tương\) \(tự\) \(với\) : \(\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\) \(và\) \(\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\) \(đều\) \(\ge2\) \(như\) \(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge2\)

\(\Rightarrow\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\ge9\) \(hay\) \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)

Bài 1:

uses crt;

var n,i,s:integer;

begin

clrscr;

write('Nhap n='); readln(n);

s:=0;

for i:=1 to n do 

 if i mod 6=0 then s:=s+i;

writeln(s);

readln;

end.

Bài 2: 

uses crt;

var a,b,c,ucln,i:integer;

begin

clrscr;

write('a='); readln(a);

write('b='); readln(b);

write('c='); readln(c);

while a<>b do 

  begin

if a>b then a:=a-b

else b:=b-a;

end;

ucln:=a;

while ucln<>c do 

 begin

if ucln>c then ucln:=ucln-c

else c:=c-ucln;

end;

writeln(ucln);

readln;

end.

Scratch hay java

\(P=a+\sqrt{\dfrac{a}{2}.2b}+\sqrt[3]{\dfrac{a}{4}.b.4c}\)

\(P\le a+\dfrac{1}{2}\left(\dfrac{a}{2}+2b\right)+\dfrac{1}{3}\left(\dfrac{a}{4}+b+4c\right)\)

\(P\le\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{16}{21};\dfrac{4}{21};\dfrac{1}{21}\right)\)