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\(\left(a\sqrt{b+1}+b\sqrt{a+1}\right)^2\le\left(a^2+b^2\right)\left(a+b+2\right)=a+b+2\le\sqrt{2\left(a^2+b^2\right)}+2=2+\sqrt{2}\)
\(\Rightarrow a\sqrt{b+1}+b\sqrt{a+1}\le\sqrt{2+\sqrt{2}}\)
\(VP^2=\frac{a+\sqrt{a^2-b}}{2}+\frac{a-\sqrt{a^2-b}}{2}+2\sqrt{\frac{\left(a+\sqrt{a^2-b}\right)\left(a-\sqrt{a^2-b}\right)}{2.2}}\)
\(=a+\sqrt{a^2-\left(a^2-b\right)}=a+\sqrt{b}=VP^2\)
Áp dụng bđt bunhiacopski cho 3 số ta có
\(\left(a\sqrt{1-b^2}+b\sqrt{1-c^2}+c\sqrt{1-a^2}\right)^2\le\left(a^2+b^2+c^2\right)\left(1-a^2+1-b^2+1-c^2\right)\Leftrightarrow\frac{9}{4}\le\left(a^2+b^2+c^2\right)\left[3-\left(a^2+b^2+c^2\right)\right]\)(1)
Đặt \(a^2+b^2+c^2=k\)
Vậy (1)\(\Leftrightarrow\frac{9}{4}\le k\left(3-k\right)\Leftrightarrow\frac{9}{4}\le3k-k^2\Leftrightarrow k^2-3k+\frac{9}{4}\le0\Leftrightarrow\left(k-\frac{3}{2}\right)^2\le0\)
Vì \(\left(k-\frac{3}{2}\right)^2\ge0\)
Suy ra \(\left(k-\frac{3}{2}\right)^2=0\Leftrightarrow k-\frac{3}{2}=0\Leftrightarrow k=\frac{3}{2}\)
Vậy \(a^2+b^2+c^2=\frac{3}{2}\)
Áp dụng BĐT cô -si \(\left(ab\le\frac{\left(a+b\right)^2}{4}\right)\) ta có :
\(\frac{1}{2}\cdot2\sqrt{ab}\left(a+b\right)\le\frac{1}{2}\cdot\frac{\left(a+b+2\sqrt{ab}\right)^2}{4}=\frac{1}{2}\cdot\frac{\left(\sqrt{a}+\sqrt{b}\right)^4}{4}=\frac{1}{8}\)
<=> \(\sqrt{ab}\left(a+b\right)\le\frac{1}{8}\)
<=> \(ab\left(a+b\right)^2\le\frac{1}{64}\)
Dấu '' = '' xảy ra khi a = b = \(\frac{1}{4}\)
BPT <=> \(\sqrt{ab}\left(a+b\right)\le\frac{1}{8}\)
\(\frac{1}{2}\cdot2\sqrt{ab}\left(a+b\right)\le\frac{1}{2}\cdot\frac{\left(a+2\sqrt{ab}+b\right)^2}{4}=\frac{1}{2}\cdot\frac{\left(\sqrt{a}+\sqrt{b}\right)^4}{4}=\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}\)
\(a\sqrt{1+a}+b\sqrt{1+b}\le\sqrt{\left(a^2+b^2\right)\left(1+a+1+b\right)}\)
\(=\sqrt{2+a+b}\le\sqrt{2+\sqrt{2\left(a^2+b^2\right)}}=\sqrt{2+\sqrt{2}}\)
Dấu = xảy ra khi \(a=b=\dfrac{1}{\sqrt{2}}\)