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b, Ta có
\(\frac{\sqrt{x}+1}{y+1}=\frac{\left(\sqrt{x}+1\right)\left(y+1\right)-y-y\sqrt{x}}{y+1}=\sqrt{x}+1-\frac{y\left(\sqrt{x}+1\right)}{y+1}\)
Mà \(y+1\ge2\sqrt{y}\)
=> \(\frac{\sqrt{x}+1}{y+1}\ge\sqrt{x}+1-\frac{1}{2}\sqrt{y}\left(\sqrt{x}+1\right)\)
Khi đó
\(P\ge\frac{1}{2}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+3-\frac{1}{2}\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)\)
Mà \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}=3\)
=> \(P\ge\frac{1}{2}.3+3-\frac{3}{2}=3\)
Vậy MinP=3 khi x=y=z=1
a/ \(\frac{2x+1}{\sqrt{x^2+2}}+\left(x+1\right)\left(\sqrt{1+\frac{2x+1}{x^2+2}}-1\right)+2x+1=0\)
\(\Leftrightarrow\frac{2x+1}{\sqrt{x^2+2}}+\frac{\left(x+1\right)\left(2x+1\right)}{\sqrt{1+\frac{2x+1}{x^2+2}}+1}+2x+1=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\frac{1}{\sqrt{x^2+2}}+\frac{x+1}{\sqrt{1+\frac{2x+1}{x^2+2}}+1}+1\right)=0\)
\(\Rightarrow x=-\frac{1}{2}\)
b/ \(Q\ge\frac{\left(x+y+z\right)^2}{xyz\left(x+y+z\right)}+\frac{\left(x^3+y^3+z^3\right)^2}{xy+yz+zx}\ge\frac{x+y+z}{xyz}+\frac{\left(x^2+y^2+z^2\right)^3}{\left(x+y+z\right)^2}\)
\(Q\ge\frac{27\left(x+y+z\right)}{\left(x+y+z\right)^3}+\frac{\left(x+y+z\right)^6}{27\left(x+y+z\right)^2}=\frac{27}{\left(x+y+z\right)^2}+\frac{\left(x+y+z\right)^4}{27}\)
\(Q\ge\frac{27}{64\left(x+y+z\right)^2}+\frac{27}{64\left(x+y+z\right)^2}+\frac{\left(x+y+z\right)^4}{27}+\frac{837}{32\left(x+y+z\right)^2}\)
\(Q\ge3\sqrt[3]{\frac{27^2\left(x+y+z\right)^4}{64^2.27\left(x+y+z\right)^4}}+\frac{837}{32.\left(\frac{3}{2}\right)^2}=\frac{195}{16}\)
"=" \(\Leftrightarrow x=y=z=\frac{1}{2}\)
Nguyễn Trúc Giang, Duy Khang, Vũ Minh Tuấn, Võ Hồng Phúc, tth, No choice teen, Phạm Lan Hương,
Nguyễn Lê Phước Thịnh, @Nguyễn Việt Lâm, @Akai Haruma
giúp em vs ạ! Cần trước 5h chiều nay ạ
Thanks nhiều
Đặt \(\sqrt{x^2+y^2}=c;\sqrt{y^2+z^2}=a;\sqrt{z^2+x^2}=b\)
Ta có:
\(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
\(\ge\frac{x^2}{\sqrt{2\left(y^2+z^2\right)}}+\frac{y^2}{\sqrt{2\left(z^2+x^2\right)}}+\frac{z^2}{\sqrt{2\left(x^2+y^2\right)}}\)
\(=\frac{1}{2\sqrt{2}}\left(\frac{c^2+b^2-a^2}{a}+\frac{a^2+c^2-b^2}{b}+\frac{b^2+a^2-c^2}{c}\right)\)
\(\ge\frac{1}{2\sqrt{2}}\left(\frac{\left(2a+2b+2c\right)^2}{2\left(a+b+c\right)}-2018\right)=\frac{1009}{\sqrt{2}}\)
Câu 1 chuyên phan bội châu
câu c hà nội
câu g khoa học tự nhiên
câu b am-gm dựa vào hằng đẳng thử rồi đặt ẩn phụ
câu f đặt \(a=\frac{2m}{n+p};b=\frac{2n}{p+m};c=\frac{2p}{m+n}\)
Gà như mình mấy câu còn lại ko bt nha ! để bạn tth_pro full cho nhé !
Câu c quen thuộc, chém trước:
Ta có BĐT phụ: \(\frac{x^3}{x^3+\left(y+z\right)^3}\ge\frac{x^4}{\left(x^2+y^2+z^2\right)^2}\) \((\ast)\)
Hay là: \(\frac{1}{x^3+\left(y+z\right)^3}\ge\frac{x}{\left(x^2+y^2+z^2\right)^2}\)
Có: \(8(y^2+z^2) \Big[(x^2 +y^2 +z^2)^2 -x\left\{x^3 +(y+z)^3 \right\}\Big]\)
\(= \left( 4\,x{y}^{2}+4\,x{z}^{2}-{y}^{3}-3\,{y}^{2}z-3\,y{z}^{2}-{z}^{3 } \right) ^{2}+ \left( 7\,{y}^{4}+8\,{y}^{3}z+18\,{y}^{2}{z}^{2}+8\,{z }^{3}y+7\,{z}^{4} \right) \left( y-z \right) ^{2} \)
Từ đó BĐT \((\ast)\) là đúng. Do đó: \(\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\frac{x^2}{x^2+y^2+z^2}\)
\(\therefore VT=\sum\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\sum\frac{x^2}{x^2+y^2+z^2}=1\)
Done.
\(M^2=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{2xy}{\sqrt{yz}}+\frac{2yz}{\sqrt{zx}}+\frac{2xz}{\sqrt{yz}}=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{2x\sqrt{y}}{\sqrt{z}}+\frac{2y\sqrt{z}}{\sqrt{x}}+\frac{2z\sqrt{x}}{\sqrt{y}}\)
Áp dụng bđt Cô-si: \(\frac{x^2}{y}+\frac{x\sqrt{y}}{\sqrt{z}}+\frac{x\sqrt{y}}{\sqrt{z}}+z\ge4\sqrt[4]{\frac{x^2}{y}.\frac{x\sqrt{y}}{\sqrt{z}}.\frac{x\sqrt{y}}{\sqrt{z}}.z}=4x\)
tương tự \(\frac{y^2}{z}+\frac{y\sqrt{z}}{\sqrt{x}}+\frac{y\sqrt{z}}{\sqrt{x}}+x\ge4y\);\(\frac{z^2}{x}+\frac{z\sqrt{x}}{\sqrt{y}}+\frac{z\sqrt{x}}{\sqrt{y}}+y\ge4z\)
=>\(M^2+x+y+z\ge4\left(x+y+z\right)\Rightarrow M^2\ge3\left(x+y+z\right)\ge3.12=36\Rightarrow M\ge6\)
Dấu "=" xảy ra khi x=y=z=4
Vậy minM=6 khi x=y=z=4
Từ giả thiết \(xy+yz+zx=5\)
ta có \(x^2+5=x^2+xy+yz+zx=\left(x+y\right)\left(z+x\right)\)
Áp dụng BĐT AM-GM , ta có
\(\sqrt{6\left(x^2+5\right)}=\sqrt{6\left(x+y\right)\left(z+x\right)}\le\frac{3\left(x+y\right)+2\left(z+x\right)}{2}=\frac{5x+3y+2z}{2}\)
CM tương tự ta được \(\sqrt{6\left(y^2+5\right)}\le\frac{3x+5y+2z}{2};\sqrt{z^2+5}\le\frac{x+y+2z}{2}\)
Cộng zế zới zế BĐt trên ta đc
\(\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}\le\frac{9x+9y+6z}{2}\)
\(=>P=\frac{3x+3y+2z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{x^2+5}}\ge\frac{2\left(3x+3y+2z\right)}{9x+9y+6z}=\frac{2}{3}\)
=> \(GTNN\left(P\right)=\frac{2}{3}khi\left(x=y=1;z=2\right)\)
Ta có \(\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}=\sqrt{6\left(x+y\right)\left(x+z\right)}+\sqrt{6\left(y+z\right)\left(y+x\right)}\)\(+\sqrt{6\left(z+x\right)\left(z+y\right)}\)
\(\le\frac{3\left(x+y\right)+2\left(x+z\right)}{2}+\frac{3\left(x+y\right)+2\left(y+z\right)}{2}+\frac{\left(z+x\right)+\left(z+y\right)}{2}\le\frac{9x+9y+6z}{2}=\frac{3}{2}\)\(\left(3x+3y+2z\right)\)
\(\Rightarrow P=\frac{3x+3y+2z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}}\ge\frac{2}{3}\)
dấu "=" xảy ra \(\Leftrightarrow x=y=1;z=2\)
Vậy \(P_{min}=\frac{2}{3}\Leftrightarrow x=y=1;z=2\)