Exercice AA-003-A

  • Math&Maroc

    a,b,ca,b,c sont des réels positifs ou nuls tels que a2+b2+c2=2(ab+bc+ca)a^2+b^2+c^2=2(ab+bc+ca) et a+b+c=2\sqrt{a}+\sqrt{b}+\sqrt{c}=2. Montrer que l'un au moins des nombres a,b,ca,b,c est égal à 1.

  • Sum sqrt(a) = 2 => sum(a) + 2sum sqrt(ab)=4
    =>(a+b+c)^2 =(4_2sum sqrt(ab) )^2
    =>1+sqrt(abc)=sum sqrt ab
    Now we have to prove thar A=(a-1)(b-1)(c-1)=0
    <=> sum(a) _ sum(ab) + abc -1 =0
    <=> 4-2(sum(sqrt(ab)) -(sum(ab))+abc-1=0
    <=> (sqrt(abc)+1_(sum(sqrt(ab)) (sqrt(abc)+3 + (sum(sqrt(ab)) =0
    Which is true (dapres la relation qu'ona demontre)

  • Since a,b,ca,b,c play symmetric roles we can suppose WLOGWLOG that abc a \geq b\geq c
    Now the first condition becomes:
    a2+b2+c2=2(ab+bc+ca)(abc)2=4bca^2+b^2+c^2=2(ab+bc+ca) \Leftrightarrow (a-b-c)^2=4bc
    abc=2bc \Leftrightarrow a-b-c=2\sqrt{bc}
    a=(b+c)2\Leftrightarrow a=(\sqrt{b}+\sqrt{c})^2
    a=b+c\Leftrightarrow \sqrt{a}=\sqrt{b}+\sqrt{c}
    The second condition gives us 2a=22\sqrt{a}=2 , therefore a=1a=1

  • @Ahmed Quand tu as enleve le carré c est soit abc=2bc a-b-c= 2\sqrt bc soit b+ca=2bc b+c-a= 2\sqrt bc
    Et le deuxieme cas te donnera comme tu as suppose que bc b\geq c que c est b=1\sqrt b =1

  • Ah merci J'avais oublie le deuxieme cas.

  • a=2(a)2=4a+2ab=4\sum\sqrt {a}=2\Longrightarrow\left(\sum \sqrt {a}\right)^2 = 4 \Longrightarrow \sum a+2 \sum \sqrt {ab}=4
     a=42aba2+2ab=16+4(ab+2a×bc)16ab\Longrightarrow \sum a =4 - 2\sum \sqrt {ab}\Longrightarrow \sum a^2 +2 \sum ab=16+4 \left(\sum ab+2 \sum a\times\sqrt {bc}\right) -16 \sum \sqrt {ab}
    4ab=16+4(ab+2(a×bc))16ab2+(a×bc)2ab=0\Longrightarrow 4\sum ab=16+4 \left(\sum ab+2 (\sum a\times \sqrt {bc})\right) -16 \sum \sqrt {ab}\Longrightarrow 2+\sum(a\times \sqrt{bc})-2\sum \sqrt{ab}=0
    2+abc×(a)2(ab)=02+2abc2(ab)=0\Longrightarrow 2+\sqrt {abc}\times(\sum \sqrt {a}) -2(\sum \sqrt {ab})=0\Longrightarrow 2+2\sqrt {abc}-2(\sum \sqrt{ab})=0
    1+abc(ab)=01+ab×(c1)c×(2c)=0\Longrightarrow 1+\sqrt {abc} - (\sum \sqrt {ab})=0\Longrightarrow 1+\sqrt{ab}\times (\sqrt {c}-1) -\sqrt {c}\times (2-\sqrt{c})=0
    (c1)×(ab+c1)=0c=1\Longrightarrow (\sqrt {c}-1)\times (\sqrt {ab}+\sqrt {c}-1) =0\Longrightarrow c=1 ou ab+c=1.\sqrt {ab}+\sqrt {c}=1.
     Donc au moins l' un des nombres a,b,ca,b, c est égal à 1.

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