Áp dụng cô si ,ta có

\(a^2+b^2\ge2ab\)

\(c^2+b^2\ge2bc\)

\(a^2+c^2\ge2ac\)

\(\Rightarrow2a^2+2b^2+2c^2\ge2ab+2ac+2bc\)

\(a^2+b^2+c^2\ge ab+ac+bc\)

\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc\ge3ab+3ac+3bc\)

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)

\(\Rightarrow200^2\ge3\left(ab+bc+ac\right)\)

\(\Rightarrow\frac{40000}{3}\ge ab+bc+ac\)

Dấu = xảy ra khi a=b=c=200/3

Ta có:  \(n^{200}<5^{300}\)=> \(n^{2\cdot100}<5^{3\cdot100}=>\left(n^2\right)^{100}<\left(5^3\right)^{100}\Leftrightarrow n^2<5^3\Leftrightarrow n^2<125\)\(\Rightarrow n^2\in\left\{0;1;4;9;16;25;36;49;64;81;100;121\right\}\)

mà n >0

\(=>n\in\left\{1;2;3;4;5;6;7;8;9;10;11\right\}\)

mà n là số nguyên dương lớn nhất

=> n = 11

Vậy n =11

n^200<5^300

=>(n^2)^100<(5^3)^100

=>n^2<5^3

=>n^2<125

=>n^2 E {0;1;4;9;...;121}

mà n lớn nhất

=>n^2=121=>n=11

vậy n=11

n^200<5^300=>(n^2)^100<(5^3)^100

=>n^2<5^3=125

=>n^2 thuộc {0;4;9;...;121}

mà n lớn nhất=>n^2=121=>n=+11

mà n nguyên dương =>n=11

tick nhé

n^200<5^300=>(n^2)^100<(5^3)^10;0=>n^2<5^3=125=>n^2={0;4;9;...;121}

ma n lon nhat=>n^2=121=>n=11

tick nhe

Ta có : \(n^{200}=\left(n^2\right)^{100};5^{300}=\left(5^3\right)^{100}=125^{100}\)

Để: \(n^{200}< 5^{300}\Rightarrow\left(n^2\right)^{100}< 125^{100}\Leftrightarrow n^2< 125\)\(\Leftrightarrow n=11\)

\(n^{200}< 5^{300}\) => \(\left(n^2\right)^{100}< 125^{100}\) => \(n^2< 125\) <=> \(11^2< 125\) => \(n=11\)

ủng hộ cho tui tui giải cho

\(5^{300}=\left(5^3\right)^{100}=125^{100}\)

\(n^{200}=\left(n^2\right)^{100}\)

n=12