Min P em có thể tự tìm đơn giản bằng AM-GM

Min R cũng khá đơn giản:

Đặt \(\left(\sqrt[3]{a};\sqrt[3]{b};\sqrt[3]{c}\right)=\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}0\le x;y;z\le1\\x^3+y^3+z^3=\dfrac{9}{8}\end{matrix}\right.\)

\(R=\dfrac{1}{1+x}+\dfrac{1}{1+y}+\dfrac{1}{1+z}\ge\dfrac{9}{3+x+y+z}\ge\dfrac{9}{3+\sqrt[3]{9\left(x^3+y^3+z^3\right)}}=\dfrac{6}{2+\sqrt[3]{3}}\)

Xét \(Q=x+y+z\)

Do \(\left(x+y+z\right)^3\ge x^3+y^3+z^3=\dfrac{9}{8}\Rightarrow x+y+z\ge\sqrt[3]{\dfrac{9}{8}}>1\Rightarrow Q-1>0\)

\(x^3+y^3+z^3=\left(x+y+z\right)^3-3\left(x+y+z\right)\left(xy+yz+zx\right)+3xyz\)

\(\Rightarrow\dfrac{9}{8}=Q^3-3Q\left(xy+yz+zx\right)+3xyz\)

\(\Rightarrow\dfrac{9}{8}=Q^3-3\left(Q-1\right)\left(xy+yz+zx\right)-3\left(xy+yz+zx-xyz\right)\)

Do \(0\le x;y;z\le1\Rightarrow\left(1-x\right)\left(1-y\right)\left(1-z\right)\ge0\)

\(\Rightarrow xy+yz+zx-xyz\ge Q-1\)  (1)

\(\Rightarrow xy+yz+zx\ge xyz+Q-1\ge Q-1\) (2)

(1);(2)\(\Rightarrow\dfrac{9}{8}\le Q^3-3\left(Q-1\right)\left(Q-1\right)-3\left(Q-1\right)\)

\(\Rightarrow8Q^3-24Q^2+24Q-9\ge0\)

\(\Rightarrow\left(2Q-3\right)\left(4Q^2-6Q+3\right)\ge0\)

Do \(4Q^2-6Q+3=4\left(Q-\dfrac{3}{4}\right)^2+\dfrac{3}{4}>0;\forall Q\)

\(\Rightarrow2Q-3\ge0\Rightarrow Q\ge\dfrac{3}{2}\)

\(Q_{min}=\dfrac{3}{2}\) khi \(\left(x;y;z\right)=\left(0;1;\dfrac{1}{2}\right)\) và hoán vị hay \(\left(a;b;c\right)=\left(0;1;\dfrac{1}{8}\right)\) và hoán vị

a: Đặt \(B=\sqrt{a+\sqrt{b}}\pm\sqrt{a-\sqrt{b}}\)

\(B^2=a+\sqrt{b}+a-\sqrt{b}\pm2\sqrt{\left(a+\sqrt{b}\right)\left(a-\sqrt{b}\right)}\)

\(=2a\pm2\sqrt{a^2-b}=2\left(a\pm\sqrt{a^2-b}\right)\)

=>\(B=\sqrt{2\left(a\pm\sqrt{a^2-b}\right)}\)

b: Đặt \(A=\sqrt{\dfrac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}}\)

=>\(A^2=\dfrac{a+\sqrt{a^2-b}}{2}+\dfrac{a-\sqrt{a^2-b}}{2}\pm2\sqrt{\dfrac{a^2-\left(\sqrt{a^2-b}\right)^2}{4}}\)

\(=\dfrac{2a}{2}\pm2\cdot\dfrac{\sqrt{a^2-a^2+b}}{2}\)

\(=a\pm\sqrt{b}\)

=>\(A=\sqrt{a\pm\sqrt{b}}\)

Có \(a^4+b^4+c^4=\left(a^2\right)^2+\left(b^2\right)^2+\left(c^2\right)^2\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3}\) 

\(\ge\dfrac{\left(\dfrac{\left(a+b+c\right)^2}{3}\right)^2}{3}\)  (áp dụng 2 lần BĐT \(x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}\))

\(=\dfrac{\left(\dfrac{4^2}{3}\right)^2}{3}=\dfrac{256}{27}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\dfrac{4}{3}\)

Vậy \(minP=\dfrac{256}{27}\) khi \(a=b=c=\dfrac{4}{3}\)

Min P dễ em có thể tự tìm đơn giản bằng AM-GM

\(P=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(=\left(a^2+b^2+c^2\right)^2-2\left(ab+bc+ca\right)^2+4abc\left(a+b+c\right)\)

\(=\left(a^2+b^2+c^2\right)^2-2\left(ab+bc+ca\right)^2+16abc\)

Do \(0\le a;b;c\le3\Rightarrow\left(3-a\right)\left(3-b\right)\left(3-c\right)\ge0\)

\(\Rightarrow3\left(ab+bc+ca\right)-9\left(a+b+c\right)+27-abc\ge0\)

\(\Rightarrow ab+bc+ca\ge\dfrac{abc+9}{3}\)

\(a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)=16-2\left(ab+bc+ca\right)\)

\(\le16-\dfrac{2}{3}\left(abc+9\right)\)

Do đó:

\(P\le\left[16-\dfrac{2}{3}\left(abc+9\right)\right]^2-2\left(\dfrac{abc+9}{3}\right)^2+16abc\)

Đặt \(abc=x\Rightarrow0\le x\le\dfrac{64}{27}\)

\(P\le\left[16-\dfrac{2}{3}\left(x+9\right)\right]^2-2\left(\dfrac{x+9}{3}\right)^2+16x\)

\(P\le\dfrac{2}{9}\left(x^2-6x+369\right)\)

\(P\le\dfrac{2}{9}x\left(x-6\right)+82\)

Do \(0\le x\le\dfrac{64}{27}\Rightarrow x-6< 0\Rightarrow\dfrac{2}{9}x\left(x-6\right)\le0\)

\(\Rightarrow P\le82\)

Dấu "=" xảy ra khi \(x=0\) hay \(\left(a;b;c\right)=\left(0;1;3\right)\) và các hoán vị

\(=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}\)

\(=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\left(\sqrt{3}-1\right)}=\sqrt{2}\left(\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\right)\)

\(=\sqrt{2}\left(\dfrac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\right)\)

\(=\sqrt{2}\left(\dfrac{6}{9-3}\right)=\sqrt{2}\)

ΔMNP vuông tại M

=>\(\widehat{MNP}+\widehat{P}=90^0\)

=>\(\widehat{N}=90^0-45^0=45^0\)

Xét ΔMNP vuông tại M có \(tanP=\dfrac{MN}{MP}\)

=>\(\dfrac{10}{MP}=tan45=1\)

=>MP=10(cm)

ΔMNP vuông tại M

=>\(MN^2+MP^2=NP^2\)

=>\(NP=\sqrt{10^2+10^2}=10\sqrt{2}\left(cm\right)\)

I am down with a cold and fever, so I can’t go to school

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