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a.
\(\dfrac{x}{x+\sqrt{3x+yz}}=\dfrac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\dfrac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}\le\dfrac{x}{x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}}\)
\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Tương tự:
\(\dfrac{y}{y+\sqrt{3y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Cộng vế:
\(VT\le\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
b.
\(VP=\dfrac{4\left(a+b+c\right)}{2\sqrt{4a\left(a+3b\right)}+2\sqrt{4b\left(b+3c\right)}+2\sqrt{4c\left(c+3a\right)}}\)
\(VP\ge\dfrac{4\left(a+b+c\right)}{4a+a+3b+4b+b+3c+4c+c+3a}\)
\(VP\ge\dfrac{4\left(a+b+c\right)}{8\left(a+b+c\right)}=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)⇒ \(ay=bx;bz=cy;cx=az\)
⇒ \(\left(\sqrt{ay}-\sqrt{bx}\right)^2+\left(\sqrt{bz}-\sqrt{cy}\right)^2+\left(\sqrt{cx}-\sqrt{az}\right)^2\)\(=0\)
⇒ \(ay+az+bx+bz+cx+cy=2\left(\sqrt{aybx}+\sqrt{bzcy}+\sqrt{cxaz}\right)\)
⇒ \(ax+ay+az+bx+by+bz+cx+cy+cz=ax+by+cz+2\left(\sqrt{axby}+\sqrt{bycz}+\sqrt{czax}\right)\)
⇒ \(\left(a+b+c\right)\left(x+y+z\right)=\left(\sqrt{ax}+\sqrt{by}+\sqrt{cz}\right)^2\)
⇒ \(\sqrt{ax}+\sqrt{by}+\sqrt{cz}=\sqrt{\left(a+b+c\right)\left(x+y+z\right)}\)
Vậy ....
Đặt \(ax^3=by^3=cz^3=k\).
Khi đó ta có:
\(VT=\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{\dfrac{k}{x}+\dfrac{k}{y}+\dfrac{k}{z}}=\sqrt[3]{k\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}=\sqrt[3]{k}\).
\(VP=\sqrt[3]{\dfrac{k}{x^3}}+\sqrt[3]{\dfrac{k}{y^3}}+\sqrt[3]{\dfrac{k}{z^3}}=\sqrt[3]{k}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\sqrt[3]{k}\).
Từ đó ta có đpcm.
Ta có: ax3 = \(\dfrac{ax^2}{\dfrac{1}{x}}\)
Tương tự ta có: ax3 = by3 = cz3
hay \(\dfrac{ax^2}{\dfrac{1}{x}}=\dfrac{by^2}{\dfrac{1}{y}}=\dfrac{cz^2}{\dfrac{1}{z}}=\dfrac{ax^2+by^2+cz^2}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\) = ax2 + by2 + cz2 (T/c dãy tỉ số bằng nhau)
\(\Rightarrow\) \(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{ax^3}=\sqrt[3]{by^3}=\sqrt[3]{cz^3}\)
= \(\dfrac{\sqrt[3]{a}}{\dfrac{1}{x}}=\dfrac{\sqrt[3]{b}}{\dfrac{1}{y}}=\dfrac{\sqrt[3]{c}}{\dfrac{1}{z}}=\dfrac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\) (đpcm)
Chúc bn học tốt!
\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=k\text{ thì }a=xk;b=yk;c=zk\)
\(VT=\sqrt[3]{x^2k}+\sqrt[3]{y^2k}+\sqrt[3]{z^2k}=\sqrt[3]{k}\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\right)\)
\(VP=\sqrt[3]{k\left(x+y+z\right)\left(x+y+z\right)}=\sqrt[3]{k}\sqrt[3]{\left(x+y+z\right)^2}\)
đề sai sai
Thử nhé
Vì P là bất đẳng thức đối xứng nên dự đoán điểm rơi \(x=y=z=\dfrac{\sqrt{2021}}{3}\)
Thay vo P ta duoc \(P=4.\sqrt{2021}\)
----------------------------------------------------------
\(P=\sum\dfrac{\left(x+y\right)\sqrt{\left(y+z\right)\left(z+x\right)}}{z}\)
Cauchy-Schwarz:
\(\Rightarrow\left(y+z\right)\left(z+x\right)\ge\left(z+\sqrt{xy}\right)^2\Rightarrow\sqrt{\left(y+z\right)\left(z+x\right)}\ge z+\sqrt{xy}\)
\(\Rightarrow P\ge\sum\dfrac{\left(x+y\right)\left(z+\sqrt{xy}\right)}{z}\ge\sum\dfrac{xz+yz+x\sqrt{y}+y\sqrt{x}}{z}=\sum x+y+\dfrac{\left(x+y\right)\sqrt{xy}}{z}\ge\sum x+y+\dfrac{2xy}{z}\)
\(\Rightarrow P\ge2(x+y+z)+2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\)
Cauchy-Schwarz: \(\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\ge\left(\sqrt{\dfrac{xy}{z}.\dfrac{yz}{z}}+\sqrt{\dfrac{yz}{x}.\dfrac{zx}{y}}+\sqrt{\dfrac{zx}{y}.\dfrac{xy}{z}}\right)^2=\left(x+y+z\right)^2\)
\(\Rightarrow P\ge2(x+y+z)+2\left(x+y+z\right)=4\left(x+y+z\right)=4\sqrt{2021}\)
\("="\Leftrightarrow x=y=z=\dfrac{\sqrt{2021}}{3}\)
Lời giải:
Ta có:
\(A=\sqrt{(x+y)(y+z)(z+x)}\left(\frac{\sqrt{y+z}}{x}+\frac{\sqrt{z+x}}{y}+\frac{\sqrt{x+y}}{z}\right)\)
\(A=\frac{(y+z)\sqrt{(x+y)(x+z)}}{x}+\frac{(z+x)\sqrt{(y+z)(y+x)}}{y}+\frac{(x+y)\sqrt{(z+x)(z+y)}}{z}\)
Áp dụng BĐT Bunhiacopxky:
\((x+y)(x+z)\geq (x+\sqrt{yz})^2\) và tương tự với những biểu thức khác suy ra:
\(A\geq \frac{(y+z)(x+\sqrt{yz})}{x}+\frac{(z+x)(y+\sqrt{xz})}{y}+\frac{(x+y)(z+\sqrt{xy})}{z}\)
hay \(A\geq 2(x+y+z)+\frac{(y+z)\sqrt{yz}}{x}+\frac{(z+x)\sqrt{zx}}{y}+\frac{(x+y)\sqrt{xy}}{z}\)
hay \(A\geq 2(x+y+z)+\underbrace{\frac{yz(y+z)\sqrt{yz}+xz(x+z)\sqrt{xz}+xy(x+y)\sqrt{xy}}{xyz}}_{M}\)
Đặt \((x,y,z)=(a^2,b^2,c^2)\)
Khi đó: \(M=\frac{a^3b^3(a^2+b^2)+b^3c^3(b^2+c^2)+c^3a^3(a^2+c^2)}{a^2b^2c^2}\)
Áp dụng BĐT AM-GM:
\(a^5b^3+a^3b^5\geq 2\sqrt{a^8b^8}=2a^4b^4\)
\(b^5c^3+c^5b^3\geq 2b^4c^4\)
\(c^5a^3+a^5c^3\geq 2c^4a^4\)
\(\Rightarrow a^3b^3(a^2+b^2)+b^3c^3(b^2+c^2)+c^3a^3(c^2+a^2)\geq 2(a^4b^4+b^4c^4+c^4a^4)\) (1)
(cộng các BĐT theo vế)
Tiếp tục AM-GM:
\(a^4b^4+b^4c^4\geq 2a^2b^4c^2; b^4c^4+c^4a^4\geq 2a^2b^2c^4; c^4a^4+a^4b^4\geq 2a^4b^2c^2\)
\(\Rightarrow a^4b^4+b^4c^4+c^4a^4\geq a^2b^2c^2(a^2+b^2+c^2)\) (2)
Từ\((1); (2)\Rightarrow a^3b^3(a^2+b^2)+b^3c^3(b^2+c^2)+c^3a^3(c^2+a^2)\geq 2a^2b^2c^2(a^2+b^2+c^2)\)
\(\Rightarrow M\geq 2(a^2+b^2+c^2)=2(x+y+z)\)
Do đó: \(A\geq 2(x+y+z)+M\geq 4(x+y+z)\Leftrightarrow A\geq 4\sqrt{2}\)
Vậy \(A_{\min}=4\sqrt{2}\Leftrightarrow x=y=z=\frac{\sqrt{2}}{3}\)
Lời giải:
Ta có:
A=√(x+y)(y+z)(z+x)(√y+zx+√z+xy+√x+yz)
A=(y+z)√(x+y)(x+z)x+(z+x)√(y+z)(y+x)y+(x+y)√(z+x)(z+y)z
Áp dụng BĐT Bunhiacopxky:
(x+y)(x+z)≥(x+√yz)2 và tương tự với những biểu thức khác suy ra:
A≥(y+z)(x+√yz)x+(z+x)(y+√xz)y+(x+y)(z+√xy)z
hay A≥2(x+y+z)+(y+z)√yzx+(z+x)√zxy+(x+y)√xyz
hay A≥2(x+y+z)+yz(y+z)√yz+xz(x+z)√xz+xy(x+y)√xyxyz M
Đặt (x,y,z)=(a2,b2,c2)
Khi đó: M=a3b3(a2+b2)+b3c3(b2+c2)+c3a3(a2+c2)a2b2c2
Áp dụng BĐT AM-GM:
a5b3+a3b5≥2√a8b8=2a4b4
b5c3+c5b3≥2b4c4
c5a3+a5c3≥2c4a4
⇒a3b3(a2+b2)+b3c3(b2+c2)+c3a3(c2+a2)≥2(a4b4+b4c4+c4a4) (1)
(cộng các BĐT theo vế)
Tiếp tục AM-GM:
a4b4+b4c4≥2a2b4c2;b4c4+c4a4≥2a2b2c4;c4a4+a4b4≥2a4b2c2
⇒a4b4+b4c4+c4a4≥a2b2c2(a2+b2+c2) (2)
Từ(1);(2)⇒a3b3(a2+b2)+b3c3(b2+c2)+c3a3(c2+a2)≥2a2b2c2(a2+b2+c2)
⇒M≥2(a2+b2+c2)=2(x+y+z)
Do đó: A≥2(x+y+z)+M≥4(x+y+z)⇔A≥4√2
Vậy Amin=4√2⇔x=y=z=√23
BĐT cần chứng minh tương đương
\(VT\ge4\left(x+y+z\right)\)
\(\Leftrightarrow\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge4\left(x+y+z\right)\)
Theo BĐT Cauchy-Schwarz và AM-GM, ta có:
\(\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge\dfrac{\left(y+z\right)\left(x+\sqrt{yz}\right)}{x}=y+z+\dfrac{\left(y+z\right)\sqrt{yz}}{x}\ge y+z+\dfrac{2yz}{x}\)
Suy ra: \(\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge2\left(x+y+z\right)-2\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)\)
Mặt khác, theo AM-GM:
\(\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)^2\ge3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\ge x+y+z\)
\(\Rightarrow\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge4\left(x+y+z\right)\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\dfrac{\sqrt{2}}{3}\)
@Phương An
Áp dụng BĐT Bunhiacopxki , ta có :
\(\left(a+b+c\right)\left(x+y+z\right)\text{≥}\left(\sqrt{ax}+\sqrt{by}+\sqrt{cz}\right)^2\)
⇔ \(\left(\sqrt{a+b+c}\right)\left(\sqrt{x+y+z}\right)\text{≥}\sqrt{ax}+\sqrt{by}+\sqrt{cz}\)
\("="\text{⇔}\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
⇒ \(\left(\sqrt{a+b+c}\right)\left(\sqrt{x+y+z}\right)\text{=}\sqrt{ax}+\sqrt{by}+\sqrt{cz}\)