Ta có x2z(x2+z2)=1z−zx2+z2⩾1z−12x" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">x2z(x2+z2)=1z−zx2+z2⩾1z−12xx2z(x2+z2)=1z−zx2+z2⩾1z−12x⇒∑x2z(x2+z2)⩾∑1z−12x=12x+12y+12z" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">⇒∑x2z(x2+z2)⩾∑1z−12x=12x+12y+12z⇒∑x2z(x2+z2)⩾∑1z−12x=12x+12y+12z⇒P⩾∑(12x+2x2)=∑(14x+14x+2x2)⩾3.314x.14x.2x23=92" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">⇒P⩾∑(12x+2x2)=∑(14x+14x+2x2)⩾3.33√14x.14x.2x2=92

∑a2+ab+1a2+3ab+1≥∑a2+ab+1a2+ab+a2+b2+c2=∑a2+ab+1a2+ab+1=∑a2+ab+1(1)" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">∑a2+ab+1√a2+3ab+$c^2$≥∑a2+ab+1√a2+ab+a2+b2+c2=∑a2+ab+1√a2+ab+1=∑√a2+ab+1(1)∑a2+ab+1a2+3ab+1≥∑a2+ab+1a2+ab+a2+b2+c2=∑a2+ab+1a2+ab+1=∑a2+ab+1(1)Áp dụng bất đẳng thức Bu-nhia-cốp-xki:[a+32(a+b)+c]2≤(1+3+1)[a+34(a+b)2+c2]≤5(a2+a2+ab+b2+c2)=5(a2+ab+1)⇒a2+ab+1≥15[a+32(a+b)+c]⇒∑a2+ab+1≥15∑[a+32(a+b)+c]=5∑a(2)" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 38.397em; min-height: 0px; padding-top: 1px; padding-bottom: 1px; width: 745px; position: relative;">[a+32(a+b)+c]2≤(1+3+1)[a+34(a+b)2+c2]≤5(a2+a2+ab+b2+c2)=5(a2+ab+1)⇒√a2+ab+1≥1√5[a+32(a+b)+c]⇒∑√a2+ab+1≥1√5∑[a+32(a+b)+c]=√5∑a(2)[a+32(a+b)+c]2≤(1+3+1)[a+34(a+b)2+c2]≤5(a2+a2+ab+b2+c2)=5(a2+ab+1)⇒a2+ab+1≥15[a+32(a+b)+c]⇒∑a2+ab+1≥15∑[a+32(a+b)+c]=5∑a(2)(1)(2)⇒dpcm" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">(1)(2)⇒dpcm

∑a2+ab+1a2+3ab+1≥∑a2+ab+1a2+ab+a2+b2+c2=∑a2+ab+1a2+ab+1=∑a2+ab+1(1)" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">∑a2+ab+1√a2+3ab+c^2≥∑a2+ab+1√a2+ab+a2+b2+c2=∑a2+ab+1√a2+ab+1=∑√a2+ab+1(1)∑a2+ab+1a2+3ab+1≥∑a2+ab+1a2+ab+a2+b2+c2=∑a2+ab+1a2+ab+1=∑a2+ab+1(1)Áp dụng bất đẳng thức Bu-nhia-cốp-xki:[a+32(a+b)+c]2≤(1+3+1)[a+34(a+b)2+c2]≤5(a2+a2+ab+b2+c2)=5(a2+ab+1)⇒a2+ab+1≥15[a+32(a+b)+c]⇒∑a2+ab+1≥15∑[a+32(a+b)+c]=5∑a(2)" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 38.397em; min-height: 0px; padding-top: 1px; padding-bottom: 1px; width: 745px; position: relative;">[a+32(a+b)+c]2≤(1+3+1)[a+34(a+b)2+c2]≤5(a2+a2+ab+b2+c2)=5(a2+ab+1)⇒√a2+ab+1≥1√5[a+32(a+b)+c]⇒∑√a2+ab+1≥1√5∑[a+32(a+b)+c]=√5∑a(2)[a+32(a+b)+c]2≤(1+3+1)[a+34(a+b)2+c2]≤5(a2+a2+ab+b2+c2)=5(a2+ab+1)⇒a2+ab+1≥15[a+32(a+b)+c]⇒∑a2+ab+1≥15∑[a+32(a+b)+c]=5∑a(2)(1)(2)⇒dpcm" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">(1)(2)⇒dpcm[a+32(a+b)+c]2≤(1+3+1)[a+34(a+b)2+c2]≤5(a2+a2+ab+b2+c2)=5(a2+ab+1)⇒a2+ab+1≥15[a+32(a+b)+c]⇒∑a2+ab+1≥15∑[a+32(a+b)+c]=5∑a(2)

6. gt $\Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$ $3(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})+3\geq 2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a} +\frac{1}{b}+\frac{1}{c})=12$ $\Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\geq3$dấu "=" $\Leftrightarrow a=b=c=1$

7. gt $\Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$ $3(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})+3\geq 2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a} +\frac{1}{b}+\frac{1}{c})=12$ $\Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\geq3$dấu "=" $\Leftrightarrow a=b=c=1$

6. gt $\Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$ $3(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})+3\geq 2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} )=12$ $\Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geq 3$dấu "=" $\Leftrightarrow a=b=c=1$

6. gt $\Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$ $3(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})+3\geq 2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a} +\frac{1}{b}+\frac{1}{c})=12$ $\Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\geq3$dấu "=" $\Leftrightarrow a=b=c=1$

đặt $1-a=x;1-b=y;1-c=z$ $\Rightarrow x,y,z>0$ và $x+y+z=2$S=$(1+\frac{2}{x})(1+\frac{2}{y})(1+\frac{2}{z})=1+\frac{2}{x}+\frac{2}{y}+\frac{2}{z}+\frac{4}{xy}+\frac{4}{yz}+\frac{4}{zx}+\frac{8}{xyz}$ $\geq 1+\frac{2.9}{x+y+z}+\frac{16}{xyz} \geq 1+9+\frac{16}{(\frac{x+y+z}{3})}{3})^{3}}=64$ dấu '=" $\Leftrightarrow a=b=c=\frac{1}{3}$

đặt $1-a=x;1-b=y;1-c=z$ $\Rightarrow x,y,z>0$ và $x+y+z=2$S=$(1+\frac{2}{x})(1+\frac{2}{y})(1+\frac{2}{z})=1+\frac{2}{x}+\frac{2}{y}+\frac{2}{z}+\frac{4}{xy}+\frac{4}{yz}+\frac{4}{zx}+\frac{8}{xyz}$ $ \geq 1+\frac{18}{x+y+z}+\frac{16}{xyz}\geq 1+9+\frac{16}{(\frac{x+y+z}{3})^{3}}=64$ dấu '=" $\Leftrightarrow a=b=c=\frac{1}{3}$

đặt $1-a=x;1-b=y;1-c=z$ $\Rightarrow x,y,z>0$ và $x+y+z=2$S=$(1+\frac{2}{x})(1+\frac{2}{y})(1+\frac{2}{z})=1+\frac{2}{x}+\frac{2}{y}+\frac{2}{z}+\frac{4}{xy}+\frac{4}{yz}+\frac{4}{zx}+\frac{8}{xyz}$ $\geq 1+\frac{2.9}{x+y+z}+\frac{16}{xyz}\geq 1+9+\frac{16}{(\frac{x+y+z}{3})}{3})^{3}}=64$ dấu '=" $\Leftrightarrow a=b=c=\frac{1}{3}$

đặt $1-a=x;1-b=y;1-c=z$ $\Rightarrow x,y,z>0$ và $x+y+z=2$S=$(1+\frac{2}{x})(1+\frac{2}{y})(1+\frac{2}{z})=1+\frac{2}{x}+\frac{2}{y}+\frac{2}{z}+\frac{4}{xy}+\frac{4}{yz}+\frac{4}{zx}+\frac{8}{xyz}$ $\geq 1+\frac{2.9}{x+y+z}+\frac{16}{xyz} \geq 1+9+\frac{16}{(\frac{x+y+z}{3})}{3})^{3}}=64$ dấu '=" $\Leftrightarrow a=b=c=\frac{1}{3}$

ĐK;$x\geq-1$ pt $\Leftrightarrow (9x+12)\sqrt{6x+8}-(19x+26)\sqrt{x+1} \leq \sqrt{(x^{2}+2x-3)^{3}} +\sqrt{x^{2}+2x-3}$ (1) Đặt $\sqrt{6x+8}=a;\sqrt{x+1}=b \Rightarrow 9x+12=a^{2}+3b^{2}+1;19x+26=3a^{2}+b^{2}+1$ (1) $\Leftrightarrow (a^{2}+3b^{2}+1)a-(3a^{2}+b^{2}+1)b\leq \sqrt{(x^{2}+2x-3)^{3}}+\sqrt{x^{2}+2x-3}$ $\Leftrightarrow (a-b)^{3}+(a-b)\leq \sqrt{(x^{2}+2x-3)^{3}}+\sqrt{x^{2}+2x-3}$ $\Leftrightarrow (a-b-\sqrt{x^{2}+2x-3})(a-b+(a-b)\sqrt{x^{2}+2x-3}+x^{2}+2x-3+1)$ $\Leftrightarrow a-b\leq \sqrt{x^{2}+2x-3}$ do (....)>0 $\Leftrightarrow \sqrt{6x+8} -\sqrt{x+1}\leq \sqrt{x^{2}+2x-3}$ $\Leftrightarrow 6x+8\leq (\sqrt{x^{2}+2x-3} +\sqrt{x+1})^{2}$ $\Leftrightarrow (\sqrt{x^{2}-1}-\sqrt{x+3})(3\sqrt{x+3}+\sqrt{x^{2}-1})\geq0$ $ \Leftrightarrow x^{2}-1\geqx+3$ $ \Leftrightarrow x \frac{1-\sqrt{17}}{2}or x\geq\frac{1+\sqrt{17}}{2}$kh vs ĐK $\Rightarrow x\geq \frac{1+\sqrt{17}}{2}$

ĐK;$x\geq-1$ pt $\Leftrightarrow (9x+12)\sqrt{6x+8}-(19x+26)\sqrt{x+1} \leq \sqrt{(x^{2}+2x-3)^{3}} +\sqrt{x^{2}+2x-3}$ (1) Đặt $\sqrt{6x+8}=a;\sqrt{x+1}=b \Rightarrow 9x+12=a^{2}+3b^{2}+1;19x+26=3a^{2}+b^{2}+1$ (1) $\Leftrightarrow (a^{2}+3b^{2}+1)a-(3a^{2}+b^{2}+1)b\leq \sqrt{(x^{2}+2x-3)^{3}}+\sqrt{x^{2}+2x-3}$ $\Leftrightarrow (a-b)^{3}+(a-b)\leq \sqrt{(x^{2}+2x-3)^{3}}+\sqrt{x^{2}+2x-3}$ $\Leftrightarrow (a-b-\sqrt{x^{2}+2x-3})(a-b+(a-b)\sqrt{x^{2}+2x-3}+x^{2}+2x-3+1)$ $\Leftrightarrow a-b\leq \sqrt{x^{2}+2x-3}$ do (....)>0 $\Leftrightarrow \sqrt{6x+8} -\sqrt{x+1}\leq \sqrt{x^{2}+2x-3}$ $\Leftrightarrow 6x+8\leq (\sqrt{x^{2}+2x-3} +\sqrt{x+1})^{2}$ $\Leftrightarrow (\sqrt{x^{2}-1}-\sqrt{x+3})(3\sqrt{x+3}+\sqrt{x^{2}-1})\geq0$ $ \Leftrightarrow x^{2}-1 \geq x+3$ $ \Leftrightarrow x \leq \frac{1-\sqrt{17}}{2}or x\geq\frac{1+\sqrt{17}}{2}$kh vs ĐK $\Rightarrow x\geq \frac{1+\sqrt{17}}{2}$