Toán Lớp 10: Tìm min $P=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2abc$
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Toán Lớp 10: Tìm min $P=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2abc$
Comments ( 1 )
Cho các số thực dương $a,b,c.$ Tìm $GTNN$ của:
$P=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2abc$
Giải.
Áp dụng BĐT $Cauchy$ cho $5$ số dương $\dfrac{1}{a^2},\dfrac{1}{b^2},\dfrac{1}{c^2},abc$ và $abc$, ta có:
$P=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2abc \\ =\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+abc+abc\\ \geq 5\sqrt[5]{\dfrac{1}{a^2}.\dfrac{1}{b^2}.\dfrac{1}{c^2}.abc.abc}=5$
Dấu $”=”$ xảy ra khi $\dfrac{1}{a^2}=\dfrac{1}{b^2}=\dfrac{1}{c^2}=abc$ hay $a=b=c=1$.
Vậy $Min(P)=5$ khi $a=b=c=1.$