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Đặt: \(\left\{{}\begin{matrix}\sqrt{a}=p\\\sqrt{b}=q\\\sqrt{x}=m\\\sqrt{y}=n\end{matrix}\right.\Leftrightarrow p;q;m;n>0\)
\(bdt\Leftrightarrow pm+qn\le\sqrt{\left(p^2+q^2\right)\left(m^2+n^2\right)}\)
Áp dụng bất đẳng thức Bunyakovsky:
\(\left(p^2+q^2\right)\left(m^2+n^2\right)\ge\left(pm+qn\right)^2\Leftrightarrow\sqrt{\left(p^2+q^2\right)\left(m^2+n^2\right)}\ge pm+qn\)
Vậy bất đẳng thức cần chứng minh đúng. Dấu "=" khi: \(\dfrac{p^2}{m^2}=\dfrac{q^2}{n^2}\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}\)
Áp dụng bất đẳng thức Mincopxki:
\(\sqrt{a^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{b^2}+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{c^2}+\dfrac{1}{a^2}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)
\(=\sqrt{\left(a+b+c\right)^2+2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)
\(\ge\sqrt{\left(a+b+c\right)^2+2.\left(\dfrac{9}{a+b+c}\right)^2}\) ( Cauchy-Schwarz)
\(=\sqrt{\left(a+b+c\right)^2+\dfrac{162}{\left(a+b+c\right)^2}}=\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)
\("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
Trời vẫn ngát xanh,gió vẫn trong lành...
a) \(a^2+9b^2-4a+6a+5=0\)
\(\Rightarrow a^2+9b^2+2a+5=0\)
\(\Rightarrow\left(a+1\right)^2+9b^2+4=0\)
\(\rightarrow ptvn\)
(T nghĩ bài này vẫn là +6b đúng hơn nhé)
b) \(a^2+2b^2+2ab-2b+2=0\)
\(\left(a+b\right)^2+\left(b-1\right)^2+1=0\)
Câu a là +6b đúng k ạ?
1 cách giải khác:
\(bdt\Leftrightarrow8\left(x^3+y^3+z^3\right)\ge\left(x+y\right)^3+\left(y+z\right)^3+\left(x+z\right)^3\)
\(\Leftrightarrow8\left(x^3+y^3+z^3\right)\ge2\left(x^3+y^3+z^3\right)+xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)\)
\(\Leftrightarrow6\left(x^3+y^3+z^3\right)\ge xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)\)
\(\Leftrightarrow3\left(x+y\right)\left(x^2-xy+y^2\right)+3\left(y+z\right)\left(y^2-yz+z^2\right)+3\left(x+z\right)\left(x^2-xz+z^2\right)\ge xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)\)
\(\Leftrightarrow3\left(x+y\right)\left(x-y\right)^2+3\left(y+z\right)\left(y-z\right)^2+3\left(x+z\right)\left(x-z\right)^2=0\)
\("="\Leftrightarrow x=y=z\)
Áp dụng bđt Cauchy-Schwarz:
\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{\left(1+1\right)^2}{2p-a-b}=\dfrac{4}{c}\)
\(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{\left(1+1\right)^2}{2p-b-c}=\dfrac{4}{a}\)
\(\dfrac{1}{p-a}+\dfrac{1}{p-c}\ge\dfrac{\left(1+1\right)^2}{2p-a-c}=\dfrac{4}{b}\)
Cộng theo vế:
\(2VT\ge4VP\Leftrightarrow VT\ge2VP\Leftrightarrowđpcm\)
\("="\Leftrightarrow a=b=c\)