Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P\le\sqrt{3\left(9a+16b+9b+16c+9c+16a\right)}=\sqrt{75\left(a+b+c\right)}=15\)
\(P_{max}=15\) khi \(a=b=c=1\)
Thầy có thể viết rõ hơn chút không ạ? Em thấy còn mơ màng lắm thầy ạ
Ta có: \(\sqrt[3]{\left(a+b\right).\frac{2}{3}.\frac{2}{3}}\le\frac{a+b+\frac{4}{3}}{3}=\frac{a+b}{3}+\frac{4}{9}\)
Tương tự rồi cộng các vế của BĐT lại, ta được: \(\sqrt[3]{\frac{4}{9}}P\le\frac{2\left(a+b+c\right)}{3}+\frac{4}{3}=2\Rightarrow P\le\sqrt[3]{18}\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)
Ta có \(\sqrt{8a^2+56}=\sqrt{8\left(a^2+7\right)}=2\sqrt{2\left(a^2+ab+2bc+2ca\right)}\)
\(=2\sqrt{2\left(a+b\right)\left(a+2c\right)}\le2\left(a+b\right)+\left(a+2c\right)=3a+2b+2c\)
Tương tự \(\sqrt{8b^2+56}\le2a+3b+2c;\)\(\sqrt{4c^2+7}=\sqrt{\left(a+2c\right)\left(b+2c\right)}\le\frac{a+b+4c}{2}\)
Do vậy \(Q\ge\frac{11a+11b+12c}{3a+2b+2c+2a+3b+2c+\frac{a+b+4c}{2}}=2\)
Dấu "=" xảy ra khi và chỉ khi \(\left(a,b,c\right)=\left(1;1;\frac{3}{2}\right)\)
a) \(P=1957\)
b) \(S=19.\)
Ta có: \(\sqrt{8a^2+56}=\sqrt{8\left(a^2+7\right)}=\sqrt{8\left(a^2+ab+2ab+2ac\right)}=2\cdot\sqrt{2\left(a+b\right)\left(a+2c\right)}\)
\(\le2\left(a+b\right)+\left(a+2c\right)=3a+2b+2c\)
Tương tự\(\hept{\begin{cases}\sqrt{8b^2+56}\le2a+3b+2c\\\sqrt{4c^2+7}=\sqrt{4c^2+ab+2ac+2bc}=\sqrt{\left(a+2c\right)\left(b+2c\right)}\le\frac{a+b+4c}{2}\end{cases}}\)
=> Q>2
Dấu "=" <=> \(\hept{\begin{cases}a=b=1\\c=1,5\end{cases}}\)
\(a;b\ge-7\) \(bđt\) \(minicopxki\)
\(\Rightarrow\sqrt{a+7}+\sqrt{b+7}=\sqrt{\sqrt{a}^2+\sqrt{7}^2}+\sqrt{\sqrt{b}^2+\sqrt{7}^2}\ge\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2+28}\)
\(\Rightarrow9\ge\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2+28}\)
\(\Leftrightarrow\left(\sqrt{a}+\sqrt{b}\right)^2\le81-28=53\Rightarrow\sqrt{a}+\sqrt{b}\le\sqrt{53}\)
\(dâu"="xảy\) \(ra\Leftrightarrow a=b=13,25\)
Áp dụng BĐT AM - GM ta có:
\(4\sqrt{ab}=2\sqrt{a.4b}\le a+4b\)
\(4\sqrt{bc}=2\sqrt{b.4c}\le b+4c\)
\(4\sqrt[3]{abc}=\sqrt[3]{a.4b.16c}\le\frac{a+4b+16c}{3}\)
Cộng theo vế 3 BĐT ta được:
\(8a+3b+4\left(\sqrt{ab}+\sqrt{bc}+\sqrt[3]{abc}\right)\le\frac{28}{3}\left(a+b+c\right)\)
\(\Rightarrow P\le\frac{28\left(a+b+c\right)}{3+3\left(a+b+c\right)^2}=\frac{14}{3}-\frac{14\left(a+b+c-1\right)^2}{3\left[\left(a+b+c\right)^2+1\right]}\le\frac{14}{3}\)
\(\Rightarrow Max_P=\frac{14}{3}\)
Đẳng thức xảy ra \(\Leftrightarrow a+b+c=1\)và \(a=4b=16c\)
\(\Leftrightarrow a=\frac{16}{21};b=\frac{4}{21};c=\frac{1}{21}\)
Chắc bạn ghi nhầm căn thức thứ 2
\(A2\sqrt{2}=2\sqrt{\left(2x+4\right)\left(x^2-2x+4\right)}+2\sqrt{\left(2y+4\right)\left(y^2-2y+4\right)}+2\sqrt{\left(2z+4\right)\left(z^2-2z+4\right)}\)
\(A2\sqrt{2}\le2x+4+x^2-2x+4+2y+4+y^2-2y+4+2z+4+z^2-2z+4\)
\(A2\sqrt{2}\le x^2+y^2+z^2+24=72\)
\(A\le18\sqrt{2}\)
Dấu "=" xảy ra khi \(x=y=z=4\)
Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)
BĐT trở thành: \(\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt{x^2+z^2+2y^2}}\le\dfrac{1}{2}\)
Ta có:
\(x^2+z^2+y^2+z^2\ge\dfrac{1}{2}\left(x+z\right)^2+\dfrac{1}{2}\left(y+z\right)^2\ge\left(x+z\right)\left(y+z\right)\)
\(\Rightarrow\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)
Tương tự: \(\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}\le\dfrac{1}{2}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)
\(\dfrac{zx}{\sqrt{z^2+x^2+2y^2}}\le\dfrac{1}{2}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)\)
Cộng vế với vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{zx+yz}{x+y}+\dfrac{xy+zx}{y+z}+\dfrac{yz+xy}{z+x}\right)=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)
\(P=\sqrt{c\left(a+b\right)}+\frac{\sqrt[3]{8.8.\left(2b+3c\right)}}{4}\)
\(\le\frac{c+a+b}{2}+\frac{8+8+2b+3c}{12}=\frac{6a+8b+9c+16}{12}\le\frac{32+16}{12}=4\)