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Thỏa mãn $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ hay $a+b+c=1$ vậy bạn?
\(\sqrt{\dfrac{ab}{c+ab}}=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)
Tương tự: \(\sqrt{\dfrac{bc}{a+bc}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)\) ; \(\sqrt{\dfrac{ca}{b+ca}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)\)
Cộng vế với vế:
\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{b}{a+b}+\dfrac{a}{a+b}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Cho a, b, c, d là các chữ số thỏa mãn: ab+ca=da ab-ca=a Tìm giá trị của d.
\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tương tự:
\(\dfrac{1}{\sqrt{b^2-bc+c^2}}\le\dfrac{1}{2}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\le\dfrac{1}{2}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(P\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng BĐT Bunhiacopxki ta có:
\(\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)\)
\(=3\left(2a+2b+2c\right)=3.2\left(a+b+c\right)=6.2021=12126\)
\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{12126}\)
Dấu ''='' xảy ra khi \(a=b=c=\dfrac{2021}{3}\)
\(Q\le\sqrt{3\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\le\sqrt{6.\sqrt{3\left(a^2+b^2+c^2\right)}}=\sqrt{6\sqrt{3}}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Lại có:
\(a^2+b^2+c^2\le1\Rightarrow0\le a;b;c\le1\)
\(\Leftrightarrow a\left(a-1\right)+b\left(b-1\right)+c\left(c-1\right)\le0\)
\(\Leftrightarrow a+b+c\ge a^2+b^2+c^2=1\)
Do đó:
\(Q^2=2\left(a+b+c\right)+2\sqrt{a^2+ab+bc+ca}+2\sqrt{b^2+ab+bc+ca}+2\sqrt{c^2+ab+bc+ca}\)
\(Q^2\ge2\left(a+b+c\right)+2\sqrt{a^2}+2\sqrt{b^2}+2\sqrt{c^2}\)
\(Q^2\ge4\left(a+b+c\right)\ge4\)
\(\Rightarrow Q\ge2\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và hoán vị
\(P=\sqrt{a+b}+\sqrt{b+c}\sqrt{c+a}\)
Aps dụng Bunhia-cốpxki : \(P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)\)
\(=6\left(a+b+c\right)\)
\(=6.2021=12126\Leftrightarrow P=\sqrt{12126}\)
Vậy \(Max\left(P\right)=\sqrt{12126}\Leftrightarrow a=b=c=\dfrac{2021}{3}\)
(Refer ;-;)
\(P^2=\left(\sqrt{4a+3}+\sqrt{4b+3}+\sqrt{4c+3}\right)^2\)
\(\le\left(1^2+1^2+1^2\right)\left(4a+3+4b+3+3c+3\right)\)
\(=63\)
\(\Rightarrow P\le\sqrt{63}=3\sqrt{7}\).
Dấu \(=\)khi \(\hept{\begin{cases}4a+3=4b+3=4c+3\\a+b+c=3\end{cases}}\Leftrightarrow a=b=c=1\).
Ta có : \(a^2+ab+b^2=\left(a+b\right)^2-ab\ge\left(a+b\right)^2-\frac{\left(a+b\right)^2}{4}=\frac{3\left(a+b\right)^2}{4}\)
\(\Rightarrow\sqrt{a^2+ab+b^2}\ge\frac{\sqrt{3}\left(a+b\right)}{2}\)
Tương tự : \(\sqrt{b^2+bc+c^2}\ge\frac{\sqrt{3}\left(b+c\right)}{2}\) ; \(\sqrt{c^2+ac+a^2}\ge\frac{\sqrt{3}\left(c+a\right)}{2}\)
Suy ra : \(\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ac+a^2}\ge\frac{\sqrt{3}}{2}.2.\left(a+b+c\right)=\sqrt{3}\)
Vậy MIN B = \(\sqrt{3}\) \(\Leftrightarrow\begin{cases}a+b+c=1\\a=b=c\end{cases}\)
\(\Leftrightarrow a=b=c=\frac{1}{3}\)
b)
https://hoc24.vn/cau-hoi/c-voi-a-b-c-la-cac-so-duong-thoa-man-dieu-kien-a-b-c-2-tim-max-q-sqrt2abcsqrt2bcasqrt2cab.8298826302
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Áp dụng Bunhiacopxki :
\(A^2=\left(\sqrt{a+b}.1+\sqrt{b+c}.1+\sqrt{c+a}.1\right)^2\le\left(a+b+b+c+c+a\right)\left(1^2+1^2+1^2\right).\)
\(\Rightarrow A^2\le6\left(a+b+c\right)=6.1=6\)
\(\Rightarrow A\le\sqrt{6}.\)
Vậy giá trị lớn nhất của \(A=\sqrt{6}\)\(\Leftrightarrow\hept{\begin{cases}a+b+c=1\\a=b=c\end{cases}\Leftrightarrow a=b=c=\frac{1}{3}}\)