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Lời giải với kiến thức lớp 8:
\(a^{2017}+b^{2017}\le a^{2018}+b^{2018}\)
\(\Leftrightarrow a^{2017}\left(a-1\right)+b^{2017}\left(b-1\right)\ge0\)
\(\Leftrightarrow a^{2017}\left(a-\frac{a+b}{2}\right)+b^{2017}\left(b-\frac{a+b}{2}\right)\ge0\)
\(\Leftrightarrow a^{2017}\cdot\frac{a-b}{2}+b^{2017}\cdot\frac{b-a}{2}\ge0\)
\(\Leftrightarrow\left(a^{2017}-b^{2017}\right)\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^{2016}+a^{2015}b+a^{2014}b^2+...+b^{2016}\right)\ge0\)
Bất đẳng thức cuối đúng với mọi a, b. Do đó bất đẳng thức đã cho là đúng.
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 \(a^2+b^2+c^2=\frac{3}{2}\)
Ta có BĐt cầnd chứng minh \(\Leftrightarrow\frac{\left(a+b\right)^2}{a^2+4}\le\frac{3}{2}\Leftrightarrow2\left(a+b\right)^2\le3\left(a^2+4\right)\)
<=>\(2\left(a^2+b^2+2ab\right)\le3\left(a^2+4\right)\Leftrightarrow2\left(4+2ab\right)\le12+3a^2\)
<=>\(4ab\le3a^2+4=4a^2+b^2\)
<=>\(0\le4a^2+b^2-4ab\Leftrightarrow0\le\left(2a-b\right)^2\left(LĐ\right)\)
=> BĐt cần chứng minh luôn đúng
^_^
a) điều kiện xác định : \(a>2;a\ne11\)
ta có : \(P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{11-a}\right):\left(\dfrac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\dfrac{1}{\sqrt{a-2}}\right)\)
\(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{\left(3+\sqrt{a-2}\right)\left(3-\sqrt{a-2}\right)}\right):\left(\dfrac{3\sqrt{a-2}+1}{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}-\dfrac{1}{\sqrt{a-2}}\right)\) \(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}\left(3-\sqrt{a-2}\right)+a+7}{\left(3+\sqrt{a-2}\right)\left(3-\sqrt{a-2}\right)}\right):\left(\dfrac{3\sqrt{a-2}+1-\sqrt{a-2}+3}{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{3\left(\sqrt{a-2}+3\right)}{\left(3+\sqrt{a-2}\right)\left(3-\sqrt{a-2}\right)}\right):\left(\dfrac{2\sqrt{a-2}+4}{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{3}{\left(3-\sqrt{a-2}\right)}\right)\left(\dfrac{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}{2\left(\sqrt{a-2}+2\right)}\right)\) \(\Leftrightarrow P=\dfrac{-\sqrt{a-2}}{2}\)
ta có : \(a+b=\sqrt{2017-a^2}+\sqrt{2017-b^2}\)
\(\Leftrightarrow\left(a+b\right)\left(\sqrt{2017-a^2}-\sqrt{2017-b^2}\right)=b^2-a^2\)
\(\Leftrightarrow b-a=\sqrt{2017-a^2}-\sqrt{2017-b^2}\)
\(\Leftrightarrow2b=2\sqrt{2017-a^2}\Leftrightarrow b^2=2017-a^2\Rightarrow\left(đpcm\right)\)
Đặt \(\hept{\begin{cases}\sqrt{a^2+b^2}=z\\\sqrt{b^2+c^2}=x\\\sqrt{c^2+a^2}=y\end{cases}}\Rightarrow\hept{\begin{cases}a=\frac{y^2+z^2-x^2}{2}\\b=\frac{x^2+z^2-y^2}{2}\\c=\frac{x^2+y^2-z^2}{2}\end{cases}}\)\(\forall\hept{\begin{cases}x,y,z>0\\x+y+z=\sqrt{2017}\end{cases}}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(b+c\le\sqrt{2\left(b^2+c^2\right)}=2x\Rightarrow\frac{a^2}{b+c}\ge\frac{y^2+z^2-x^2}{2\sqrt{2}x}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(2\sqrt{2}\cdot VT\ge\frac{y^2+z^2-x^2}{x}+\frac{y^2+x^2-z^2}{z}+\frac{x^2+z^2-y^2}{y}\)
\(=\frac{y^2}{x}+\frac{z^2}{x}+\frac{y^2}{z}+\frac{x^2}{z}+\frac{x^2}{y}+\frac{z^2}{y}-\left(x+y+z\right)\)
\(\ge\frac{\left(2\left(x+y+z\right)\right)^2}{2\left(x+y+z\right)}-\sqrt{2017}=\sqrt{2017}\)
\(\Rightarrow2\sqrt{2}\cdot VT\ge\sqrt{2017}\Rightarrow VT\ge\frac{\sqrt{2017}}{2\sqrt{2}}=VP\)