Lời giải:
$P=\frac{a^2b^2+b^2c^2+c^2a^2}{abc}$

Áp dụng BĐT AM-GM, dạng $(x+y+z)^2\geq 3(xy+yz+xz)$ ta có:

$(a^2b^2+b^2c^2+c^2a^2)^2\geq 3(a^2b^4c^2+a^4b^2c^2+a^2b^2c^4)$

$=3a^2b^2c^2(a^2+b^2+c^2)=3a^2b^2c^2$

$\Rightarrow a^2b^2+b^2c^2+c^2a^2\geq \sqrt{3}abc$

$\Rightarrow P=\frac{a^2b^2+b^2c^2+c^2a^2}{abc}\geq \sqrt{3}$

Vậy $P_{\min}=\sqrt{3}$. Giá trị này đạt tại $a=b=c=\frac{1}{\sqrt{3}}$

Đặt \(P=\dfrac{a^3}{a^2+b^2+ab}+\dfrac{b^3}{b^2+c^2+bc}+\dfrac{c^3}{c^2+a^2+ca}\)

Ta có: \(\dfrac{a^3}{a^2+b^2+ab}=a-\dfrac{ab\left(a+b\right)}{a^2+b^2+ab}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^3b^3}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)

Tương tự: \(\dfrac{b^3}{b^2+c^2+bc}\ge\dfrac{2b-c}{3}\) ; \(\dfrac{c^3}{c^2+a^2+ca}\ge\dfrac{2c-a}{3}\)

Cộng vế:

\(P\ge\dfrac{a+b+c}{3}=673\)

Dấu "=" xảy ra khi \(a=b=c=673\)

#include <bits/stdc++.h>

using namespace std;

long long a[1000],i,n,ln,t,k,nn;

int main()

{

cin>>n;

for (i=1; i<=n; i++) cin>>a[i];

ln=LLONG_MIN;

for (i=1; i<=n; i++) ln=max(ln,a[i]);

cout<<"So lon nhat la: "<<ln<<endl;

cout<<"VI tri la: ";

for (i=1; i<=n; i++) if (ln==a[i]) cout<<i<<" ";

cout<<endl;

t=0;

for (i=1; i<=n; i++)

if (a[i]>0) t+=a[i];

cout<<"Tong cac so duong la: "<<t<<endl;

cin>>k;

for (i=1; i<=n; i++)

if (a[i]%k==0) cout<<a[i]<<" ";

cout<<endl;

nn=LLONG_MAX;

for (i=1; i<=n; i++)

nn=min(nn,a[i]);

cout<<nn;

return 0;

}

nộp rồi mà

  là số chẵn

uses crt;

var a:array[1..250]of integer;

n,i,t,max,min:integer;

begin

clrscr;

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

for i:=1 to n do 

  begin

write('A[',i,']='); readln(a[i]);

end;

t:=0;

for i:=1 to n do 

 if a[i] mod 3=0 then t:=t+a[i];

writeln('Tong cac so la boi cua 3 la: ',t);

max:=a[1];

min:=a[1];

for i:=1 to n do 

begin

if max<a[i] then max:=a[i];

if min>a[i] then min:=a[i];

end;

writeln('Gia tri lon nhat la: ',max);

writeln('Gia tri nho nhat la: ',min);

readln;

end.